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Stream: deprecated: mathematics

Topic: homotopy fibers

John Baez (Apr 27 2023 at 15:50):

I'm going to try something new here: I'll talk about some stuff I'm thinking about, and see if anyone has questions or has something to say. If you don't understand what I'm saying, please ask questions! If you know stuff about this topic, I'd like to hear it!

(On other forums I'm getting disappointed with how few people talk about my posts.)

Given a functor $F : A \to X$, any object $x \in X$ has a "homotopy fiber", where an object is an object $a \in A$ together with an isomorphism $\alpha: F(a) \to x$, and a morphism is 'the obvious thing' (explanation available on request).

John Baez (Apr 27 2023 at 15:51):

This is the obvious sort of categorification of when you have a function between sets, any point in the codomain has a 'fiber' or 'inverse image' consisting of all points in the domain that map to it.

John Baez (Apr 27 2023 at 15:54):

I guess more technically this homotopy fiber is the 'isocomma object' of $F: A \to X$ and $G: 1 \to X$, where $G$ is the functor that picks out the object $X$.

John Baez (Apr 27 2023 at 15:55):

I'm interested in certain examples these days. Generalizing from these examples we get this:

John Baez (Apr 27 2023 at 15:57):

Suppose you have algebras $A$ and $B$ over some field and a homomorphism $f: A \to B$. Then you get adjoint functors between their categories modules $A\mathsf{Mod}$ and $B\mathsf{Mod}$: 'restriction along $f$'

$f^\ast : B\mathsf{Mod} \to A\mathsf{Mod}$

and its left adjoint which I guess might be called 'extension along $f$'.

$f_* : A\mathsf{Mod} \to B\mathsf{Mod}$

John Baez (Apr 27 2023 at 16:09):

This sends any $A$ module $M$ to the $B$-module $B \otimes_A M$ where $B$ becomes a right $A$-module using $f: A \to B$.

John Baez (Apr 27 2023 at 16:19):

So what I'm interested in are the homotopy fibers of $f^\ast$ and $f_\ast$ in some really concrete examples.

John Baez (Apr 27 2023 at 16:21):

The homotopy fiber of $f^*$ over some $A$-module $M$ is intuitively "the category of ways of extending the $A$-module structure on $M$ to a $B$-module structure".

John Baez (Apr 27 2023 at 16:24):

The homotopy fiber of $f_\ast$ is intuitively "the category of ways you can get the $B$-module $M$ from extending some $A$-module".

Morgan Rogers (he/him) (Apr 27 2023 at 16:30):

I'm following so far, but isn't there another adjoint that you could add to this picture? Is restriction along f also cocontinuous?

Morgan Rogers (he/him) (Apr 27 2023 at 16:31):

(Specifically, I'm expecting $\mathrm{Hom}_A(M,B)$ to provide the right adjoint)

Jean-Baptiste Vienney (Apr 27 2023 at 16:56):

I'm interested by the "explanation available on request" on what is an homotopy fiber.

Mike Shulman (Apr 27 2023 at 17:10):

John Baez said:

The homotopy fiber of $f_*$ over some $A$-module $M$ is intuitively "the set of ways of extending the $A$-module structure on $M$ to a $B$-module structure".

The homotopy fiber of $f^\ast$ is intuitively "the set of ways you can get the $B$-module $M$ from extending some $A$-module".

Aren't those reversed?

Mike Shulman (Apr 27 2023 at 17:13):

As a topologist and topos theorist, I'm used to $f_*$ denoting a right adjoint of $f^*$, which as Morgan mentioned also exists here ("coextension of scalars"); I would write $f_!$ for the left adjoint of $f^*$. I know that algebraic geometers have different conventions; is $f_*$ commonly used for extension of scalars?

Mike Shulman (Apr 27 2023 at 17:15):

Anyway, as for your question, the obvious first class of examples that occurs to me is when $A=k$ is the field itself. Then the $f^*$-fiber over a vector space $M$ is the set of $B$-module structures on it.

John Baez (Apr 27 2023 at 17:42):

Mike Shulman said:

John Baez said:

The homotopy fiber of $f_*$ over some $A$-module $M$ is intuitively "the set of ways of extending the $A$-module structure on $M$ to a $B$-module structure".

The homotopy fiber of $f^\ast$ is intuitively "the set of ways you can get the $B$-module $M$ from extending some $A$-module".

Aren't those reversed?

Yes, fixed.

John Baez (Apr 27 2023 at 17:48):

Mike Shulman said:

As a topologist and topos theorist, I'm used to $f_*$ denoting a right adjoint of $f^*$, which as Morgan mentioned also exists here ("coextension of scalars"); I would write $f_!$ for the left adjoint of $f^*$. I know that algebraic geometers have different conventions; is $f_*$ commonly used for extension of scalars?

I don't know what the hell different people do with all this $f^\ast, f_\ast, f_!$ notation - I find it incredibly confusing. Different people seem to do different things. I'm using $f^\ast$ here because an algebra homomorphism $f: A \to B$ is giving rise to a functor going "backwards", $f^\ast : A\mathsf{Mod} \to B \mathsf{Mod}$.

(So, we're getting a contravariant functor from some category of algebras to some category of some very nice categories, and when we get everything running really smoothly, perhaps adding some side conditions if necessary, it becomes a contravariant equivalence of bicategories.)

John Baez (Apr 27 2023 at 17:52):

Morgan Rogers (he/him) said:

I'm following so far, but isn't there another adjoint that you could add to this picture? Is restriction along f also cocontinuous?

There probably is another adjoint quite generally, but in fact I'm working with semisimple algebras, whose categories of (finite-dimensional) modules are semisimple categories, and this seems to ensure that the that my left adjoint $f_\ast$ is also a right adjoint to $f^\ast$. So in my particular context I've got an [[ambidextrous adjunction]], which is very fun but makes it unnecessary to roam the land looking for further adjoints.

John Baez (Apr 27 2023 at 17:55):

I should add that I'm interested in really specific features of really specific examples, so I haven't put much effort into improving my overall understanding of the general theory.

John Baez (Apr 27 2023 at 17:56):

Let me mention an example and pose some puzzles about it. I'm looking at algebras over $\mathbb{R}$. My go-to example of a homomorphism of algebras $f: A \to B$ is the inclusion $i: \mathbb{R} \to \mathbb{C}$.

John Baez (Apr 27 2023 at 17:58):

So, $f^\ast$ gives me the "underlying real vector space of a complex vector space", while $f_\ast$ gives me the "complexification of a real vector space":

$f_\ast(V) = \mathbb{C} \otimes_{\mathbb{R}} V$

John Baez (Apr 27 2023 at 17:59):

$f^\ast(W) = \{ \text{real-linear maps } \mathbb{R} \to W \}$

John Baez (Apr 27 2023 at 18:03):

Here's the puzzle:

1. Given a real vector space $V$, what is the set of isomorphism classes in the homotopy fiber of $f^\ast$ over $V$?

2. Given a complex vector space $W$, what is the set of isomorphism classes in the homotopy fiber of $f_\ast$ over $W$?

John Baez (Apr 27 2023 at 18:04):

I'm really just asking for any other, less category-ridden and I hope more easily understood, description of these two sets.

John Baez (Apr 27 2023 at 18:10):

Both these sets are famous (in certain quarters anyway - famous enough to have Wikipedia pages), but they're also fairly easy to describe without knowing any of the known stuff.

John Baez (Apr 27 2023 at 18:13):

Jean-Baptiste Vienney said:

I'm interested by the "explanation available on request" on what is an homotopy fiber.

The short answer is: the homotopy fiber of a functor $F: A \to X$ over a point $x$ is the category of ways you can get $x$ (up to isomorphism) by applying the functor $F$ to some object of $A$. The "up to isomorphism" part is what makes it a homotopy fiber instead of an ordinary fiber.

In homotopy theory we are often interested in a map $F: A \to X$ between topological spaces, and then the homotopy fiber of a point $x \in X$ consists of points $a \in A$ that don't need to have $F(a) = x$ "on the nose": instead, it's enough to have a continuous path from $F(a)$ to $x$.

John Baez (Apr 27 2023 at 23:50):

I hope you see that here we are using an analogy between

• isomorphisms between objects in categories

and

• paths between points in spaces.

This analogy is ultimately much more than an analogy, especially if we stick to categories that are just groupoids, or $n$-categories that are $n$-groupoids.

Peter Arndt (Apr 28 2023 at 09:44):

For puzzle 2, the homotopy fiber of of the complexification functor, I think the answer is given by the theory of Galois descent. This would say that the isomorphism classes of the homotopy fiber correspond to conjugations on the complex vector space $V$, i.e. maps $r\colon V \to V$ such that $r(z \cdot v)=\bar{z}\cdot r(v)$ and $r(v+v')=r(v)+r(v')$.

Peter Arndt (Apr 28 2023 at 09:48):

The homotopy fiber is always non-empty: I can just choose a basis of $V$, take its $\mathbb{R}$-linear span in $V$. That gives me an $\mathbb{R}$-vector space $W$ and the inclusion map induces an isomorphism $F(W)\to V$.

Peter Arndt (Apr 28 2023 at 09:50):

An isomorphism in the homotopy fiber would be an isomorphism $f$ of $\mathbb{R}$-vector spaces such that the triangle with $F(f)$ on top and $V$ at the bottom commutes.

Peter Arndt (Apr 28 2023 at 09:52):

The point is that this triangle is a diagram of complex vector spaces and $\mathbb{C}$-linear maps.

Peter Arndt (Apr 28 2023 at 09:56):

So it seems that an isomorphism class in the fiber over $V$ is given by a choice of $\mathbb{R}$-subvector space of $V$ that $\mathbb{C}$-spans $V$ (this is also called an $\mathbb{R}$-form of $V$).

Peter Arndt (Apr 28 2023 at 10:00):

That is already a more down to earth formulation than saying "homtopy fiber". Theorem 2.14 in the notes linked above says that the $\mathbb{R}$-forms of $V$ in turn correspond to conjugations. This correspondence is also pretty concrete.

John Baez (Apr 28 2023 at 15:55):

Great answer, @Peter Arndt. I think we don't need to know Galois descent ahead of time to answer question 2). We can just figure out the answer from scratch. But it is indeed the sort of question Galois descent is designed to answer.

Let me answer question 1), which has a somewhat different flavor.

John Baez (Apr 28 2023 at 15:59):

Let's think about what an object in a homotopy fiber of the forgetful functor

$f^\ast: \mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

actually is. We start with a real vector space $V \in \mathsf{Vect}_{\mathbb{R}}$.

Intuitively, an object in the homotopy fiber over $V$ is a way of making $V$ into a complex vector space.

But following the definition, an object in the homotopy fiber over $V$ is acually a complex vector space $W$ whose underlying real vector space $f^\ast(W)$ is equipped with an isomorphism to $V$, say

$\alpha: f^\ast(W) \to V$

John Baez (Apr 28 2023 at 16:04):

We need to reconcile our intuition with the definition!

Since $W$ is a complex vector space, its underlying real vector space $f^\ast(W)$ comes with a complex structure: a real-linear endomorphism $J$ with $J^2 = -1$. This is just multiplication by $i$ on $W$, as seen in the world of real vector spaces.

John Baez (Apr 28 2023 at 16:05):

We can use the isomorphism $\alpha$ to transport this complex structure from $f^*(W)$ to $V$.

It's easy to use the complex structure on $V$ to promote this real vector space to a complex vector space.

So indeed, an object in the homotopy fiber gives a way to make $V$ into a complex vector space! But it seems to give more: this other vector space $W$, and this isomorphism $\alpha$.

To reconcile our intuition with the definition, we need to check that this extra stuff is just fluff.

"Fluff" is stuff that goes away when you switch to working in an equivalent category.

John Baez (Apr 28 2023 at 16:08):

So, we should suspect that the homotopy fiber of $f^\ast$ over $V$ is equivalent to the category where:

• objects are complex structures $J$ on $V$.
• the only morphisms are identity morphisms.

John Baez (Apr 28 2023 at 16:20):

We can do this by messing around with commutative diagrams - and importantly, the definition of morphism in the homotopy fiber, which I haven't used yet!

John Baez (Apr 28 2023 at 16:23):

The set of complex structures on a real vector space $V$ can be described even more explicitly.

John Baez (Apr 28 2023 at 16:23):

It's empty if $V$ is odd-dimensional.

John Baez (Apr 28 2023 at 16:24):

So say $V$ is even-dimensional. We might as well say $V = \mathbb{R}^{2n}$.

John Baez (Apr 28 2023 at 16:27):

The group $GL(2n,\mathbb{R})$ acts transitively on the set of complex structures, since for each $J: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ with $J^2$ we can find a basis for $\mathbb{R}^{2n}$ in which $J$ is block diagonal with $2 \times 2$ blocks of this form:

$\left( \begin{array}{cr} 0 & -1 \\ 1 & 0 \end{array} \right)$

John Baez (Apr 28 2023 at 16:29):

The stabilizer of the "standard" complex structure on $\mathbb{R}^{2n} \cong \mathbb{C}^n$ is, by definition, the group of complex-linear transformations $GL(n,\mathbb{C})$.

John Baez (Apr 28 2023 at 16:30):

So, the set of complex structures on $\mathbb{R}^{2n}$ is isomorphic to

$GL(2n,\mathbb{R})/GL(n,\mathbb{C})$

John Baez (Apr 28 2023 at 16:31):

So that's our homotopy fiber - a very concrete answer to puzzle 1!

John Baez (Apr 28 2023 at 16:33):

Note that in this case it's a discrete category, a mere set. I think that's because $f^\ast$ is faithful - it's only forgetting structure, not stuff.

John Baez (Apr 28 2023 at 16:39):

Using the insights we'd gain by carefully working through this example, we can become quite efficient at working out homotopy fibers, at least in similar cases. Frankly I just think

"All complex structures on $\mathbb{R}^{2n}$ are the same up to a linear transformation of $\mathbb{R}^{2n}$, so $GL(2n,\mathbb{R})$ acts transitively on the set of these complex structures, and the stabilizer is $GL(n,\mathbb{C})$, so the homotopy fiber is $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$."

John Baez (Apr 28 2023 at 16:40):

This brings up another puzzle:

Puzzle 3: find an answer to Puzzle 2 in which we describe the homotopy fiber of "complexification"

$f_\ast : \mathsf{Vect}_\mathbb{R} \to \mathsf{Vect}_\mathbb{C}$

as the quotient of some group by a subgroup.

David Egolf (Apr 28 2023 at 17:30):

By the way, in case it's helpful for anyone else, I found this resource helpful for understanding a bit more of the above example: Kobayashi S., Nomizu K. - Foundations of differential geometry. Vol.2., chapter 9, section 1 (pages 114-116).

Mike Shulman (Apr 28 2023 at 17:36):

John Baez said:

Note that in this case it's a discrete category, a mere set. I think that's because $f^\ast$ is faithful - it's only forgetting structure, not stuff.

Faithful, and also conservative. In general, the homotopy fibers of a faithful functor are (equivalent to) posets. For instance, if $U$ is the forgetful functor from topological spaces to sets, then the homotopy fiber of $U$ over a set $X$ is the poset of topologies on $X$, ordered by refinement. But when the functor is also conservative, these posets are equivalent to discrete sets.

John Baez (Apr 28 2023 at 18:01):

Thanks! I always forget what a [[conservative functor]] is, because I say such a functor "reflects isomorphisms".

John Baez (Apr 28 2023 at 18:02):

So I was just about to pose two more abstract puzzles, based on the example I just worked through - for people who like category theory more than concrete examples from algebra and geometry. (I've heard such people exist.)

John Baez (Apr 28 2023 at 18:06):

Puzzle 4: Given a functor $F: \mathsf{A} \to \mathsf{X}$, find simple conditions under which all its homotopy fibers are sets - that is, discrete categories.

In this case we can really talk about the set of ways an object in $\mathsf{X}$ comes from an object in $\mathsf{A}$.

Puzzle 5: Give further conditions under which the homotopy fiber of an object $x \in \mathsf{X}$ is, not only a set, but a set that's a quotient of easily described groups $G/H$ .

John Baez (Apr 28 2023 at 18:07):

But Mike answered Puzzle 4 better than I could. So let me fill in some details of his answer.

John Baez (Apr 28 2023 at 18:08):

Say we have two objects in the homotopy fiber over $x \in \mathsf{X}$, for example

$a \in \mathsf{A} , \alpha: F(a) \to x$

and

$a' \in \mathsf{A} , \alpha': F(a') \to x$

John Baez (Apr 28 2023 at 18:09):

I never said what a morphism between them was. Well, I said it was the "obvious thing" - but what's that? Clearly it should start out being a morphism

$f: a \to a'$

John Baez (Apr 28 2023 at 18:09):

This then gives a morphism

$F(f) : F(a) \to F(a')$

John Baez (Apr 28 2023 at 18:11):

And this then has the opportunity to form a commutative triangle with

$\alpha: F(a) \to x$

and

$\alpha': F(a') \to x$

So, clearly we should demand that it does:

$\alpha' \circ F(f) = \alpha$

John Baez (Apr 28 2023 at 18:12):

And so that's what a morphism in the homotopy fiber is: a morphism $f: a \to a'$ obeying $\alpha' \circ F(f) = \alpha$.

John Baez (Apr 28 2023 at 18:15):

Now suppose $F$ is faithful. Let's try to show the homotopy fiber of $F$ is a preorder, i.e. there's at most one morphism $f$ between objects in the homotopy fiber.

John Baez (Apr 28 2023 at 18:15):

We can always solve the above equation for $F(f)$:

$F(f) = \alpha'^{-1} \circ \alpha$

John Baez (Apr 28 2023 at 18:17):

So we know what $F(f)$ must be. But if $F$ is faithful this means we know what $f: a \to a'$ must be!

John Baez (Apr 28 2023 at 18:20):

So we've shown

$F$ faithful $\implies$ the homotopy fibers of $F$ are preorders.

Puzzle 6. Is the converse true?

John Baez (Apr 28 2023 at 18:20):

(I haven't thought about this but it's an obvious question.)

John Baez (Apr 28 2023 at 18:22):

Next suppose that $F$ "reflects isomorphisms", i.e. is "conservative": both these terms mean that

$F(f)$ is an isomorphism $\implies$ $f$ is an isomorphism.

John Baez (Apr 28 2023 at 18:24):

Not even using those conditions, whenever we have a morphism $f$ in the homotopy fiber over $f$ we always get

$F(f) = \alpha'^{-1} \circ \alpha$

so $F(f)$ must be an isomorphism.

John Baez (Apr 28 2023 at 18:24):

But when $F$ reflect isomorphism this means that $f$ is an isomorphism!

John Baez (Apr 28 2023 at 18:25):

So we've shown

$F$ reflects isomorphisms $\implies$ the homotopy fibers of $F$ are groupoids.

John Baez (Apr 28 2023 at 18:28):

This raises yet another puzzle I haven't thought about:

Puzzle 7: Is the converse true?

So now, if $F$ is faithful and it reflects isomorphisms, the homotopy fibers of $F$ are preorders and groupoids. This means that if there's a morphism between two objects in a homotopy fiber it's unique and it's an isomorphism. This implies that the homotopy fiber is equivalent to a discrete category!

John Baez (Apr 28 2023 at 18:29):

So we've shown

$F$ is faithful and reflects isomorphisms $\implies$ the homotopy fibers of $F$ are equivalent to discrete categories.

John Baez (Apr 28 2023 at 18:30):

or less pedantically

$F$ is faithful and reflects isomorphisms $\implies$ the homotopy fibers of $F$ are sets.

where I'm using "are" in the modern, hip mathematician's sense of the verb "to be".

John Baez (Apr 28 2023 at 18:35):

So, I've pounded Puzzle 4 to death, and the pounding process made two more puzzles shoot out!

Martti Karvonen (Apr 28 2023 at 19:13):

John Baez said:

So we've shown

$F$ faithful $\implies$ the homotopy fibers of $F$ are preorders.

Puzzle 6. Is the converse true?

The converse isn't true. Consider the category $\bullet\rightrightarrows\bullet$ and let $F$ map both of the nontrivial morphisms to the unique map $\{0,1\}\to\{0\}$ in Set. Any homotopy fiber is now a preorder, but the functor is not faithful.

Mike Shulman (Apr 28 2023 at 19:16):

However, the converse is true if $F$ is a fibration or an opfibration.

John Baez (Apr 28 2023 at 20:43):

Nice observations!

To me, the most important unsolved puzzle is Puzzle 5, which to me is the key to understanding a large class of homotopy fibers very concretely as [[homogeneous spaces]].

Puzzle 5 might be hard to understand if you didn't follow my example where I showed the homotopy fiber of the forgetful functor $\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$ over $\R^{2n}$ is $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$.

John Baez (Apr 28 2023 at 20:44):

So, if anyone is having trouble, just ask questions.

Mike Shulman (Apr 28 2023 at 22:15):

So for any functor $F:A\to X$ and $x\in X$, the homotopy fiber over $x$ has an action by the group ${\rm Aut}_D(x)$. Namely, given an isomorphism $F(a) \cong x$, we can compose it with any automorphism of $x$ to get another isomorphism $F(a) \cong x$. Thus, if $F$ is faithful and conservative, so that this homotopy fiber is equivalent to a set, it is moreover an ${\rm Aut}_D(x)$-set. Hence, like any set with a group action, it is a disjoint union of orbits, each of which is a homogeneous space. And if the action happens to be transitive, then it is a single orbit, i.e. a homogeneous space.

Transitivity of the action means directly that if we have $\phi : F(a) \cong x$ and $\phi' : F(a') \cong x$, there is $\psi : a\cong a'$ and $\chi : x\cong x$ making a commutative square. But then $\chi$ is uniquely determined by the other three data, and if this holds for all $x$ then the hypothesis might as well be just that $F(a)\cong F(a')$. So, if $F$ is faithful and conservative, then its homotopy fibers are homogeneous spaces if it is injective on isomorphism classes.

Mike Shulman (Apr 28 2023 at 22:16):

The homogeneous space in question should then be ${\rm Aut}_D(x) / {\rm Aut}_A(a)$, for some $a\in A$ such that $F(a) \cong x$.

John Baez (Apr 28 2023 at 22:28):

:tada: YES! :tada:

So the extra condition we need to get this to work, beside $F$ being faithful and conservative, is that it "creates isomorphisms". This is jargon for

$F(a) \cong F(a') \implies a \cong a'$

or in other words what Mike said: $F$ is injective on isomorphism classes.

John Baez (Apr 28 2023 at 22:29):

All three conditions hold in the case of our forgetful functor

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

John Baez (Apr 28 2023 at 22:31):

It's faithful: two complex-linear maps between complex vector spaces are equal if the corresponding real-linear maps between the underlying real vector spaces are equal.

John Baez (Apr 28 2023 at 22:32):

It's conservative: if the underlying real-linear map of a complex-linear map is an iso, the original complex-linear map was an iso.

John Baez (Apr 28 2023 at 22:32):

And it creates isomorphisms: if the underlying real vector spaces of two complex vector spaces are isomorphic (= have the same dimension), then the same is true of the original complex vector spaces!

Mike Shulman (Apr 28 2023 at 22:33):

"Injective on isomorphism classes" feels like a weird condition, category-theoretically.

John Baez (Apr 28 2023 at 22:46):

True! I've been bumping into it lately. It's probably worth checking out what happens to our homotopy fiber when that third condition is not true. We should still get something nice.

But anyway, first I just want to tell the lurking students that in this example, using your result, the homotopy fiber of $\mathbb{R}^{2n}$ turns out to be

$\mathrm{Aut}_{\mathsf{Vect}_\mathbb{R}} (\mathbb{R}^{2n}) / \mathrm{Aut}_{\mathsf{Vect}_\mathbb{C}} (\mathbb{C}^{n})$

which is better known as

$\mathrm{GL}(2n,\mathbb{R}) / \mathrm{GL}(n,\mathbb{C})$

John Baez (Apr 28 2023 at 22:48):

And this is a famous space: the space of complex structures on $\mathbb{R}^n$. We can think of it as the space of linear maps

$\{ J : \mathbb{R}^n \to \mathbb{R}^n \vert J^2 = -1 \}$

John Baez (Apr 28 2023 at 22:50):

Now, a while back @Peter Arndt answered Puzzle 2, which was: what are the homotopy fibers of the left adjoint to the forgetful functor I've just been talking about! This left adjoint is "complexification"

$f_\ast: \mathsf{Vect}_{\mathbb{R}} \to \mathsf{Vect}_{\mathbb{C}}$

(using my notation, which the topos theorists dislike). It sends any real vector space $V$ to $\mathbb{C} \otimes_{\mathbb{R}} V$.

John Baez (Apr 28 2023 at 22:52):

@Peter Arndt showed that the homotopy fiber over $\mathbb{C}^n$ of this left adjoint is the set of $n$-dimensional real subspaces $V \subseteq \mathbb{C}^n$ with the property that $V + iV = \mathbb{C}^n$.

John Baez (Apr 28 2023 at 22:56):

But it would be nice if we could describe this, too, as a quotient of groups. So:

Puzzle 8: Is complexification faithful? Is it conservative? Is it injective on isomorphism classes?

Mike Shulman (Apr 28 2023 at 22:58):

John Baez said:

It's probably worth checking out what happens when that condition is not true. We should still get something nice.

I guess what we get is the disjoint union of homogeneous spaces $\coprod_a {\rm Aut}_X(x) / {\rm Aut}_A(a)$, where $a$ ranges over one object in each isomorphism class whose image is the isomorphism class of $x$.

John Baez (Apr 28 2023 at 22:59):

That's what I bet too! It's sort of a slightly groupoidy analogue of a homogeneous space.

John Baez (Apr 28 2023 at 23:00):

But the "numerator" is the same for each summand; only the denominator varies.

John Baez (Apr 28 2023 at 23:03):

I have some important examples where this comes up, coming from representation theory. So far my examples here concern the real representations of the algebras $\mathbb{R} \hookrightarrow \mathbb{C}$. But if we change its square root of -1 to a square root of +1, $\mathbb{C}$ would turn into $\mathbb{R} \oplus \mathbb{R}$. If we copied our story using the inclusion $\mathbb{R} \hookrightarrow \mathbb{R} \oplus \mathbb{R}$, we would get nonisomorphic $\mathbb{R} \oplus \mathbb{R}$ that have isomorphic underlying real vector spaces.

Mike Shulman (Apr 28 2023 at 23:05):

John Baez said:

That's what I bet too! It's sort of a slightly groupoidy analogue of a homogeneous space. But the "numerator" is the same for each summand; only the denominator varies.

Also known as a not-necessarily-transitive $G$-set...

David Egolf (Apr 28 2023 at 23:26):

This is interesting! I'm working on understanding some of the things said above. Since questions are welcome, I thought I would ask for some explanation/clarification.

I'll begin by stating what I think I understand so far.

• Let $f^*: \mathsf{Vect}_\mathbb{C} \to \mathsf{Vect}_\mathbb{R}$ be our forgetful functor of interest, sending each complex vector space to its underlying real vector space.
• Let $W$ be a complex vector space. Then $f^*(W)$ has a "complex structure" on it. That is, it has an endomorphism $J$ with $J^2= -1_{f^*(W)}$. We know this because the real-linear map that sends $v$ to $iv$ for any $v$ in $f^*(W)$ will provide such an endomorphism.
• Let $U$ be some real vector space. We are interested in understanding the set of complex structures it has. We assume it has even dimension so say $U = \mathbb{R}^{2n}$. (Odd dimension vector spaces don't have a complex structure. To see that, we can take the determinant of each side of $J^2=-1_U$, and note the right-hand side is only positive when the dimension of $U$ is even.)
• Let us assume that $W$ is such that $f^*(W)$ is isomorphic to $U$ in $\mathsf{Vect}_\mathbb{R}$ , so there exists an isomorphism $\alpha: f^*(W) \to U$. I think we can equivalently say that $W$ together with $\alpha$ is an object in the homotopy fiber of $U$ with respect to our forgetful functor $f^*$.

David Egolf (Apr 28 2023 at 23:30):

Now I'll move to the things that seem less clear to me.

Somewhere along the line, "actions" come into play, and I don't quite understand how yet.

On possibly a related note, let $J$ be the "canonical" complex structure on $f^*(W)$. This is the complex structure that multiples vectors by $i$. For any automorphism $s: U \to U$, we can produce a new isomorphism from $f^*(W)$ to $U$ (by composing after $\alpha$). (I think that means that the automorphisms of $U$ induce a map on the objects of the fiber of $U$). Given any isomorphism $\beta$ from $f^*(W)$ to $U$, we can get a complex structure on $U$ as $\beta \circ J \circ \beta^{-1}$ (I think!).

So, each automorphism of $U$ - and these automorphisms are elements of $GL(2n,\mathbb{R})$ - induces a complex structure on $U$. And I think this happens for each complex vector space $W$ such that $f^*(W)$ is isomorphic to $U$.

Somehow quotient groups come into this, which I am not following yet. Maybe the idea is that different element of $GL(2n,\mathbb{R})$ can induce the same complex structure on $U$, and so we want to just pick one representative from $GL(2n,\mathbb{R})$ for each unique resulting complex structure?

Any corrections or clarifications would be most welcome! I think it's slowly starting to make sense, but I haven't put all the pieces together yet.

John Baez (Apr 29 2023 at 01:01):

David Egolf said:

Now I'll move to the things that seem less clear to me.

Somewhere along the line, "actions" come into play, and I don't quite understand how yet.

Where? I don't remember saying that word. Maybe Peter Arndt used it when he invoked the theory of Galois descent?

"Action" means a lot of closely related things in math.

Remember that a function $f: S \to T$ can act on an element $s \in S$ to give the element $f(s) \in T$.

A monoid, e.g. a group, $G$ can act on an object of a category: see [[action]].

And so on.

David Egolf (Apr 29 2023 at 01:23):

John Baez said:

The group $GL(2n,\mathbb{R})$ acts transitively on the set of complex structures, since for each $J: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ with $J^2$ we can find a basis for $\mathbb{R}^{2n}$ in which $J$ is block diagonal with $2 \times 2$ blocks of this form:

$\left( \begin{array}{cr} 0 & -1 \\ 1 & 0 \end{array} \right)$

This is the mention of actions I was thinking of. (Unless I'm wrong to use the word "action" here to refer to the acting of the group mentioned?) Mike Shulman talks about actions as well, above, and I suspect the two are related!

I think the idea is that the automorphisms of a real vector space form a group, and this acts on the objects of the homotopy fibre category associated with it (?).

Maybe it would be helpful if I could come up with a more specific question. I think I need to take a break for now, though. (But this thread has been great fun to think about!)

John Baez (Apr 29 2023 at 04:40):

Okay, you're right: I mentioned that some group is acting on a set, and this is the traditional concept of 'group action':

• Wikipedia, Group action.

In an algebra course you might learn about groups abstractly, but a group that's not acting on anything is like a tool that's not being used yet. The main thing groups do is act on things! I hope you've seen some version of this idea, it's the idea of 'symmetry'. The Wikipedia article gives dozens of examples.

The modern general version of this concept of action is that a group $G$ gives a category $BG$ with one object $\ast$ and morphisms corresponding to elements of $G$. Given any category $C$ with an object $c \in C$, a functor $F: BG \to C$ with $F(\ast) = c$ is called an action of $G$ on the object $c$.

The idea here is that $F$ sends elements $g \in G$ to isomorphisms $F(g) : c \to c$, with

$F(gh) = F(g) F(h)$

$F(1) = 1_c$

John Baez (Apr 29 2023 at 04:40):

So, elements of the group are turning into 'symmetries' of the object $c$.

John Baez (Apr 29 2023 at 04:45):

Note that the functor $1: BG \to BG$ makes the group $G$ act on the one object $\ast \in BG$. Here the object is just a stupid thing that has no personality except that it's being acted on by $G$!

John Baez (Apr 29 2023 at 04:46):

Its whole mission in life is to be acted on by $G$.

John Baez (Apr 29 2023 at 05:02):

Note also that any object $c$ in any category has a group $\mathrm{Aut}(c)$ whose elements are automorphisms, which are isomorphisms $f: c \to c$. This group always acts on $c$!

David Egolf (Apr 29 2023 at 05:35):

I think that all makes sense! Thanks for spelling it out. (I have run across this concept now and then, before).

I'm still hazy on the exact role that the action mentioned above plays in helping us characterize a homotopy fiber. When I have more energy, I would like to ask a more specific question along those lines. (Assuming you're still up for more questions! Thanks as always for the time and effort that you spend on this zulip.)

Nathanael Arkor (Apr 29 2023 at 08:13):

John Baez said:

All three conditions hold in the case of our forgetful functor

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

Is it also pseudomonic? This would imply the other conditions, but is a more natural categorical property.

Martti Karvonen (Apr 29 2023 at 13:49):

John Baez said:

So we've shown

$F$ reflects isomorphisms $\implies$ the homotopy fibers of $F$ are groupoids.

This raises yet another puzzle I haven't thought about:

Puzzle 6: Is the converse true?

This should probably be puzzle 7. Anyway, this time the converse is true: given $f\colon A\to B$ for which $F(f)$ is an isomorphism, $f$ induces a morphism in the homotopy fiber of $F(B)$ and is thus invertible (both in the fiber and in the original cat).

John Baez (Apr 29 2023 at 14:19):

David Egolf said:

I think that all makes sense! Thanks for spelling it out. (I have run across this concept now and then, before).

Since I'm in love with symmetry and things with lots of symmetry:

truncated icosidodecahedron

I tend to regard the study of group actions as the entry-point to the most beautiful part of the universe.

I'm still hazy on the exact role that the action mentioned above plays in helping us characterize a homotopy fiber.

Sure! The key thing is that if a group $G$ acts transitively on a set $X$ (for any any two points $x, y \in X$ there exists a $g\in G$ so that $g x=y$, we can describe the set $X$ very explicitly, in a useful way!

When I have more energy, I would like to ask a more specific question along those lines.

Sure!

(Assuming you're still up for more questions! Thanks as always for the time and effort that you spend on this zulip.)

Thanks! I like talking about math and physics, especially when there's a group of people asking each other probing questions, remembering the answers, and continuing to dig deeper - there's something very pleasurable about it.

John Baez (Apr 29 2023 at 14:23):

Nathanael Arkor said:

John Baez said:

All three conditions hold in the case of our forgetful functor

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

Is it also pseudomonic? This would imply the other conditions, but is a more natural categorical property.

I always forget what [[pseudomonic functors]] are concretely, though the abstract definition makes sense. Okay, they're faithful, and full on isomorphisms.

No, that functor

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

is very much not full on isomorphisms: not every real-linear automorphism of $\mathbb{C}$ (or its underlying real vector space, which we could call $\mathbb{R}^2$) is a complex-linear automorphism!

John Baez (Apr 29 2023 at 14:25):

The group of real-linear automorphisms is called $GL(2,\mathbb{R})$ - it's the group of invertible 2 $\times$ 2 real matrices, and it's a manifold of dimension 3. The group of complex-linear automorphisms is called $GL(1,\mathbb{C})$ - it's the group of invertible 1 $\times$ 1 complex matrices, also known as $\mathbb{C}^\ast$, the complex numbers with zero removed. This is a manifold of dimension 2, so it's much smaller.

John Baez (Apr 29 2023 at 14:26):

Also $GL(2,\mathbb{R})$ is nonabelian, and $GL(1,\mathbb{C})$ is abelian.

John Baez (Apr 29 2023 at 14:29):

You can think of $GL(2,\mathbb{R})$ as the group of all real-linear transformations of the plane $\mathbb{R}^2$, and $GL(1,\mathbb{C})$ as the subgroup that preserves angles.

John Baez (Apr 29 2023 at 14:30):

So alas this more elegant pseudomonicity condition does not hold - but I'm not really sorry, because if it did I believe the homotopy fibers would all be 1-point sets!

John Baez (Apr 29 2023 at 14:30):

And then I wouldn't have any fun with them.

Mike Shulman (Apr 29 2023 at 14:40):

By the way, although pseudomonic includes faithful and implies injective-on-isomorphism-classes, it does not imply conservative. (Edit: oops, yes it does!)

Nathanael Arkor (Apr 29 2023 at 14:47):

Mike Shulman said:

By the way, although pseudomonic includes faithful and implies injective-on-isomorphism-classes, it does not imply conservative.

Does it not? If $F(g)$ is an isomorphism, then it is in $\mathrm{Iso}_D(F x, F y)$, and the action of $F$ on isomorphisms induces a bijection of this set with $\mathrm{Iso}_C(x, y)$, and so $g$ ought to be an isomorphism, no? (The nLab page states that pseudomonicity implies conservativity.)

David Egolf (Apr 29 2023 at 19:58):

Hooray! I feel like I understand a lot more about puzzle 5 now:

Puzzle 5: Give further conditions under which the homotopy fiber of an object x∈X is, not only a set, but a set that's a quotient of easily described groups G/H .

To help solidify my understanding, and in case it's interesting to anyone reading, let me try to sketch how the solution to puzzle 5 goes in my own words. (I don't have a more specific question at this time, unfortunately!) Any corrections/clarifications are always welcome.

We saw earlier that if our functor $F: A \to D$ is faithful and reflects isomorphisms, then the homotopy fiber of an object $x \in D$ is both a preorder and groupoid. That means that the objects of the homotopy fiber fall into equivalence classes, where there is one isomorphism between any two objects in an equivalence class, and no morphisms between objects in different equivalence classes. To describe this category pretty well, it suffices to count how many equivalence classes it has. So that's our goal: count these equivalence classes, and thereby provide a good description of the homotopy fiber of $x$.

To count these equivalence classes, we will use the orbit-stabilizer theorem. To use this, we need to define an action on these equivalence classes. Each automorphism of $x$ changes a homotopy fiber object into a new one by post-composition. And I think it turns out that if two homotopy fibers are isomorphic before this operation is applied, they are isomorphic afterwards. So that means that post-composition with an automorphism of $x$ defines a function on the equivalence classes of the objects of the homotopy fiber of $x$. Actually, I think we end up getting a group action, where the automorphisms of $x$ act on these equivalence classes. Each automorphism of $x$ induces a bijection of equivalence classes of homotopy fiber objects.

However, some of the automorphisms of $x$ will only shuffle around elements inside the equivalence classes in the homotopy fiber of $x$. Picking a particular equivalence class, we want to consider all the automorphisms of $x$ that map that equivalence class to itself. This is the stabilizer subgroup of the automorphisms of $X$ with respect to our chosen equivalence class. I think it turns out that the automorphisms of $X$ induced by the automorphisms of $F(a) \cong X$ in the image of $F$ end up forming this stabilizer subgroup. That means that the stabilizer subgroup for our chosen equivalence class has $|Aut_A(a)|$ elements.

Our acting group $Aut_D(x)$ has $|Aut_D(x)|$ elements. Using the stabilizer subgroup theorem, the orbit of a particular equivalence class of homotopy fiber objects has $|Aut_D(x)|/|Aut_A(a)|$ elements. If this action is transitive, that means that our equivalence class orbit consists of all the equivalence classes, so in this transitive case there are $|Aut_D(x)|/|Aut_A(a)|$ equivalence classes of objects in the homotopy fiber of $x$.

We are interested in knowing when this action of $Aut_D(x)$ is transitive on equivalence classes of the homotopy fiber. It suffices to investigate the conditions under which the action of $Aut_D(x)$ on the objects of the homotopy fiber is transitive up to isomorphism. We can describe this condition in terms of a square that needs to commute (see Mike Shulman's post for more details). For the square to exist, it is sufficient if the functor $F$ satisfies this condition: if $F(a) \cong F(a')$, then $a \cong a'$ for any objects $a$ and $a'$ of $A$.

Mike Shulman (Apr 29 2023 at 20:01):

Nathanael Arkor said:

Mike Shulman said:

By the way, although pseudomonic includes faithful and implies injective-on-isomorphism-classes, it does not imply conservative.

Does it not? If $F(g)$ is an isomorphism, then it is in $\mathrm{Iso}_D(F x, F y)$, and the action of $F$ on isomorphisms induces a bijection of this set with $\mathrm{Iso}_C(x, y)$, and so $g$ ought to be an isomorphism, no? (The nLab page states that pseudomonicity implies conservativity.)

Yep, my bad. Sorry, I guess my brain sputtered. Thanks.

John Baez (Apr 29 2023 at 20:03):

Nathanael Arkor said:

Mike Shulman said:

By the way, although pseudomonic includes faithful and implies injective-on-isomorphism-classes, it does not imply conservative.

Does it not? If $F(g)$ is an isomorphism, then it is in $\mathrm{Iso}_D(F x, F y)$, and the action of $F$ on isomorphisms induces a bijection of this set with $\mathrm{Iso}_C(x, y)$, and so $g$ ought to be an isomorphism, no? (The nLab page states that pseudomonicity implies conservativity.)

Suppose $F$ induces a bijection from $\mathrm{Iso}_C(x,y)$ to $\mathrm{Iso}_D(Fx,Fy)$, and let's try to show $F$ is conservative. So, given $g: x \to y$ for which $Ff$ is an isomorphism, we want to show $g$ was an isomorphism. Thanks to our bijection we know there's some isomorphism $f: x \to y$ such that $Ff = Fg$, but I don't think that implies $g$ is an isomorphism. So, we're not succeeding.

However, if $F$ is pseudomonic we also know it's faithful, so $Ff = Fg$ implies $f = g$, so in this case we do succeed.

John Baez (Apr 29 2023 at 20:09):

Puzzle 9: Suppose $F: C \to D$ is a functor between groupoids. Is it true that

$F$ is pseudomonic $\iff$ all the homotopy fibers of $F$ are (equivalent to) sets with 0 or 1 points

?

John Baez (Apr 29 2023 at 20:11):

I believe by now we've already shown the $\implies$ part. The $\iff$ statement would be nice since it's a categorification of this: if $F: S \to T$ is a functor between sets, then

$F$ is monic $\iff$ all the fibers of $F$ have 0 or 1 points.

John Baez (Apr 29 2023 at 20:12):

(Experts may now want to mumble something like "in a general topos... subterminal...".)

John Baez (Apr 29 2023 at 20:38):

David Egolf said:

Hooray! I feel like I understand a lot more about puzzle 5 now:

Great!

Puzzle 5: Give further conditions under which the homotopy fiber of an object x∈X is, not only a set, but a set that's a quotient of easily described groups G/H .

To help solidify my understanding, and in case it's interesting to anyone reading, let me try to sketch how the solution to puzzle 5 goes in my own words. (I don't have a more specific question at this time, unfortunately!) Any corrections/clarifications are always welcome.

We saw earlier that if our functor $F: A \to D$ is faithful and reflects isomorphisms, then the homotopy fiber of an object $x \in D$ is both a preorder and groupoid. That means that the objects of the homotopy fiber fall into equivalence classes, where there is one isomorphism between any two objects in an equivalence class, and no morphisms between objects in different equivalence classes.

Right! And a category with the property you just described is equivalent, as a category, to the [[discrete category]] on some set. So showoffs will breezily say that this category is equivalent to a set, or simply that it "is" a set. (If they are nice, they will make air quotes around the word "is" when saying this to beginners. Later they wind up essentially redefining the word "is".)

To describe this category pretty well, it suffices to count how many equivalence classes it has. So that's our goal: count these equivalence classes, and thereby provide a good description of the homotopy fiber of $x$.

Hmm, you are now using some sophisticated machinery called "counting", which reduces sets to numbers. But instead of counting the equivalence classes, I claim what you're actually about to do is describe a bijection between the set of equivalence classes and the quotient G/H of some interesting groups G and H.

In other words, if your set is called X, you're not just showing

$|X| = |G/H|$

where the bars denote cardinality. You're doing something better: you're describing a specific bijection

$X \cong G/H$

This is strictly better, since clearly it implies $|X| = |G/H|$, but it also does more: for example, you can use it to give "names" to elements of $X$. Using your bijection, any element of $X$ can be called the equivalence class of $[g]$ for some $g \in G$. $g$ is probably not unique, but we know exactly when $[g] = [g']$. This can be quite practical in examples.

In what follows I'll occasionally point out how you can easily do better than just showing $|X| = |G/H|$.

To count these equivalence classes, we will use the orbit-stabilizer theorem. To use this, we need to define an action on these equivalence classes. Each automorphism of $x$ changes a homotopy fiber object into a new one by post-composition. And I think it turns out that if two homotopy fibers are isomorphic before this operation is applied, they are isomorphic afterwards. So that means that post-composition with an automorphism of $x$ defines a function on the equivalence classes of the objects of the homotopy fiber of $x$. Actually, I think we end up getting a group action, where the automorphisms of $x$ act on these equivalence classes.

Yes!

Each automorphism of $x$ induces a bijection of equivalence classes of homotopy fiber objects.

However, some of the automorphisms of $x$ will only shuffle around elements inside the equivalence classes in the homotopy fiber of $x$. Picking a particular equivalence class, we want to consider all the automorphisms of $x$ that map that equivalence class to itself. This is the stabilizer subgroup of the automorphisms of $X$ with respect to our chosen equivalence class. I think it turns out that the automorphisms of $X$ induced by the automorphisms of $F(a) \cong X$ in the image of $F$ end up forming this stabilizer subgroup.

Yes!

That means that the stabilizer subgroup for our chosen equivalence class has $|Aut_A(a)|$ elements.

Yes! But you can do better simply by saying that the stabilizer subgroup "is" $Aut_A(a)$ - or more pedantically, is isomorphic to $Aut_A(a)$.

(This is doing better by doing less - an important trick in math.)

Our acting group $Aut_D(x)$ has $|Aut_D(x)|$ elements. Using the stabilizer subgroup theorem, the orbit of a particular equivalence class of homotopy fiber objects has $|Aut_D(x)|/|Aut_A(a)|$ elements. If this action is transitive, that means that our equivalence class orbit consists of all the equivalence classes, so in this transitive case there are $|Aut_D(x)|/|Aut_A(a)|$ equivalence classes of objects in the homotopy fiber of $x$.

Yes! And again, you can do better by doing less. What the stabilizer subgroup theorem really gives you is a bijection between the quotient group $Aut_D(x)/Aut_A(a)$ and the set of equivalence classes of objects in the homotopy fiber of $x$.

In other words, just leave out those vertical lines and get a more powerful result! :upside_down:

(By the way, instead of saying "equivalence classes of objects in the homotopy fiber" I'd say "isomorphism classes of objects in the homotopy fiber" since your equivalence relation is just "being isomorphic". This is the nitpicky sort of point I'd only make when grading homework, but "equivalence class" only makes sense if the reader remembers what equivalence relation you're talking about, but "isomorphism class" says exactly which equivalence relation we've got here.)

We are interested in knowing when this action of $Aut_D(x)$ is transitive on equivalence classes of the homotopy fiber. It suffices to investigate the conditions under which the action of $Aut_D(x)$ on the objects of the homotopy fiber is transitive up to isomorphism. We can describe this condition in terms of a square that needs to commute (see Mike Shulman's post for more details). For the square to exist, it is sufficient if the functor $F$ satisfies this condition: if $F(a) \cong F(a')$, then $a \cong a'$ for any objects $a$ and $a'$ of $A$.

Right! A functor obeying this extra condition is said to "create isomorphisms". More jargon... not so important.

John Baez (Apr 29 2023 at 20:47):

So, we can summarize all the work you just did in a way that sounds incredibly intimidating, as follows:

Theorem. Suppose $F: A \to D$ is faithful and it both reflects and creates isomorphisms. Then there is a bijection between the homotopy fiber of $x \in D$ and the set

$Aut_D(x) / Aut_A(a)$

where $a \in A$ is any object with $F(a) \cong x$.

John Baez (Apr 29 2023 at 20:48):

This is an example of how the journey is more fun than reaching the destination: the final result would probably seem rather dry and obscure if you hadn't gone through the process of seeing why it's true.

John Baez (Apr 29 2023 at 20:50):

But this is a fundamental result in 'Klein geometry'.

Felix Klein massively generalized Euclidean and various known non-Euclidean geometries by saying that you get different kinds of geometry from different groups. To do geometry you first choose a group G of symmetries. Then any subgroup H will preserve some type of 'geometrical figure' like a point or line, and G/H will be the set of all figures of that type.

John Baez (Apr 29 2023 at 20:51):

We are extending his idea a bit by seeing how this sort of situation shows up when you have a "forgetful functor" $F: A \to D$, and G/H is a homotopy fiber.

John Baez (Apr 29 2023 at 20:53):

I should admit that I owe most of my understanding of Klein geometry and homotopy fibers to James Dolan. However, I don't think he ever came out and stated the above Theorem.

John Baez (Apr 29 2023 at 22:14):

For anyone who wants a distilled version of what we've proved in this discussion, go here:

• nLab, [[essential fiber]].

I've added a bunch of propositions summarizing what we've done - because I need to remember this stuff.

The nLab calls it the "essential fiber", not the "homotopy fiber".

Mike Shulman (Apr 30 2023 at 00:38):

Re: puzzle 8, note that for a functor between groupoids, pseudomonic is the same as fully faithful.

Mike Shulman (Apr 30 2023 at 00:38):

Where did you get this terminology "creates isomorphisms" for "injective on isomorphism classes"? I would have expected "creates isomorphisms" to mean the same as either "conservative" or "pseudomonic".

John Baez (Apr 30 2023 at 00:43):

I got it from someone you know.

John Baez (Apr 30 2023 at 00:46):

I forget who told me that she uses this term. Previously @Todd Trimble had been calling this property "reflecting isomorphicness", which has the advantage of being clear if you already know "reflecting isomorphisms", and the disadvantage of using a somewhat ugly word (for a rather important concept).

Mike Shulman (Apr 30 2023 at 00:47):

I see that there were also objections to the terminology at that MO answer. What about "reflecting isomorphy"?

Mike Shulman (Apr 30 2023 at 00:48):

(Actually I don't see anything wrong with "injective on isomorphism classes".)

Jean-Baptiste Vienney (Apr 30 2023 at 00:54):

John Baez said:

For anyone who wants a distilled version of what we've proved in this discussion, go here:

• nLab, [[essential fiber]].

I've added a bunch of propositions summarizing what we've done - because I need to remember this stuff.

The nLab calls it the "essential fiber", not the "homotopy fiber".

Thanks, here is finally a definition of this fiber thing!! (other than "obvious" "thing" "sort of" etc. :upside_down:)

John Baez (Apr 30 2023 at 01:23):

I said they were obvious because these are the kind of details one can learn to fill in. Basically when an object of some sort is described by diagrams, you can guess what the morphisms between these objects should be. But later I explained everything:

Say we have two objects in the homotopy fiber over $x \in \mathsf{X}$, for example

$a \in \mathsf{A} , \alpha: F(a) \stackrel{\sim}{\to} x$

and

$a' \in \mathsf{A} , \alpha': F(a') \stackrel{\sim}{\to} x$

I never said what a morphism between them was. Well, I said it was the "obvious thing" - but what's that? Clearly it should start out being a morphism

$f: a \to a'$

And this then has the opportunity to form a commutative triangle with

$\alpha: F(a) \to x$

and

$\alpha': F(a') \to x$

So, clearly we should demand that it does:

$\alpha' \circ F(f) = \alpha$

And so that's what a morphism in the homotopy fiber is: a morphism $f: a \to a'$ obeying $\alpha' \circ F(f) = \alpha$.

John Baez (Apr 30 2023 at 01:25):

Mike Shulman said:

(Actually I don't see anything wrong with "injective on isomorphism classes".)

Nothing wrong with it except it's a wee bit long when you have to say it over and over, as @Todd Trimble and @Joe Moeller and I need to. We haven't finalized what we'll say though.

Nathanael Arkor (Apr 30 2023 at 08:21):

The name "essentially injective" appears sensible to me, because it's the usual definition of injectivity, but where equality has been replaced by isomorphism (analogous, say, to [[essentially surjective functor]]). (Though I don't see this suggestion in the MathOverflow thread, so maybe I'm overlooking something.)

John Baez (Apr 30 2023 at 17:56):

That term makes sense to me.

John Baez (Apr 30 2023 at 18:27):

So, maybe we're almost done, except it seems nobody has tried this one:

Puzzle 3: Describe the homotopy fiber of "complexification"

$G: \mathsf{Vect}_\mathbb{R} \to \mathsf{Vect}_\mathbb{C}$
$G: V \mapsto \mathbb{C} \otimes_\mathbb{R} V$

over $W \in \mathsf{Vect}_\mathbb{C}$ as the quotient of some group by a subgroup.

John Baez (Apr 30 2023 at 18:30):

First I guess we need to see if our Theorem applies:

John Baez (Apr 30 2023 at 18:31):

Theorem. Suppose $F: A \to X$ is faithful, reflects isomorphisms and is injective on isomorphism classes. Then there is a bijection between the homotopy fiber of $x \in X$ and the set

$Aut_X(x) / Aut_A(a)$

where $a \in A$ is any object with $F(a) \cong x$.

David Egolf (Apr 30 2023 at 21:47):

This is a very basic question, but I'm still trying to understand what $\mathbb{C} \otimes_\mathbb{R} V$ is. I thought that the $\mathbb{R}$ under the $\otimes$ meant that the two vector spaces being tensored are both real vector spaces and that the resulting vector space is also a real vector space. ("An Introduction to Homological Algebra" by Rotman says on page 76: "For example, if $V$ and $W$ are vector spaces over a field $k$, then their tensor product $V \otimes_k W$ is also a vector space over $k$.") However, $\mathbb{C} \otimes_\mathbb{R} V$ is listed as an object of $\mathsf{Vect}_\mathbb{C}$ above, so I'm confused.

John Baez (Apr 30 2023 at 22:26):

@David Egolf - You can think of $\mathbb{C} \otimes_{\mathbb{R}} V$ as a real vector space if you like, but here the whole point is that we can also think of it as a complex vector space. For this, notice that you can multiply guys in $\mathbb{C} \otimes_{\mathbb{R}} V$ by complex numbers. For example, given

$a \otimes v \in \mathbb{C} \otimes_{\mathbb{R}} V$

we can multiply it by $b \in \mathbb{C}$ and get

$ba \otimes v \in \mathbb{C} \otimes_{\mathbb{R}} V$

John Baez (Apr 30 2023 at 22:29):

You have to check that this procedure is well-defined, since guys in $\mathbb{C} \otimes_{\mathbb{R}} V$ are actually equivalence classes of formal linear combinations of guys like $a \otimes v$. And you have to check that this winds up making $\mathbb{C} \otimes_{\mathbb{R}} V$ into a complex vector space - you have to check that the laws of a complex vector space hold!

But it's all true - so you don't actually have to check it unless you want to. :upside_down:

John Baez (Apr 30 2023 at 22:31):

Using this procedure, we get

$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{R}^n \cong \mathbb{C}^n$

where this is an isomorphism of complex vector spaces. And we say that "complexifying $\mathbb{R}^n$ gives $\mathbb{C}^n$.

John Baez (Apr 30 2023 at 22:33):

So the idea is that if $V$ is a real vector space, and we suddenly pretend we can multiply guys in $V$ by complex numbers - the sort of thing physicists do without blinking an eye, since nobody has ever stopped a physicist from doing this - we are actually replacing $V$ by its complexification, $\mathbb{C} \otimes_{\mathbb{R}} V$.

John Baez (Apr 30 2023 at 22:35):

There's a whole general theory of games like this, which I am mightily resisting getting into. If you want a small taste of it, read this:

• Wikipedia, Change of rings: constructions.

John Baez (Apr 30 2023 at 22:39):

But more to the point, complexification is a functor

$\mathsf{Vect}_{\mathbb{R}} \to \mathsf{Vect}_{\mathbb{C}}$

that is left adjoint to the 'forgetful functor'

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

that foolishly forgets we can multiply by complex numbers, and thinks of a complex vector space as a mere real vector space with dimension twice as big - like thinking of $\mathbb{C}^n$ as $\mathbb{R}^{2n}$.

John Baez (Apr 30 2023 at 22:41):

You've said you haven't got the hang of adjoint functors yet, but the best way to get it is to keep seeing examples of pairs of functors that go back and forth between 2 categories: often the right adjoint involves "forgetting stuff you knew" while the left adjoint involves "puffing something up to make it better".

John Baez (Apr 30 2023 at 22:43):

Here the right adjoint

$\mathsf{Vect}_{\mathbb{C}} \to \mathsf{Vect}_{\mathbb{R}}$

forgets how to multiply by complex numbers and turns a complex vector space into a real vector space, while the left adjoint

$\mathsf{Vect}_{\mathbb{R}} \to \mathsf{Vect}_{\mathbb{C}}$

puffs up our real vector space by "going ahead and pretending we know how to multiply vectors by complex numbers".

John Baez (Apr 30 2023 at 22:44):

I happen to be busy thinking about homotopy fibers of certain interesting adjoint functors, for my next talk on the tenfold way - and this is the simplest example.

John Baez (Apr 30 2023 at 22:50):

You're making me think a lot about the textbook on category theory I might write someday. Right now you're making me realize it should say: "The left adjoint of forgetting something you know how to do is pretending you know how to do something you don't know how to do."

John Baez (Apr 30 2023 at 22:51):

You may have seen those obnoxious tee-shirts that say JUST DO IT. That's a very left adjoint thing to say.

David Egolf (Apr 30 2023 at 23:08):

Thanks for explaining, that makes a lot of sense!

John Baez (Apr 30 2023 at 23:09):

Great!

Mike Shulman (May 01 2023 at 05:05):

John Baez said:

First I guess we need to see if our Theorem applies.

I think it does, doesn't it? It takes an $n$-real-dimensional space to an $n$-complex-dimensional space, and two vector spaces are isomorphic iff they have the same dimension, so it should be essentially injective. And it should take a real-linear map given by some matrix to the complex-linear map given by the same matrix, which implies faithfulness, and also conservativity since it preserves determinants and a linear map is invertible iff it has a nonzero determinant.

That suggests that the homotopy fiber over an $n$-dimensional complex vector space is $GL(n,\mathbb{C})/GL(n,\mathbb{R})$.

John Baez (May 01 2023 at 17:11):

Yes, all that sounds right to me! And while I haven't seen $GL(n,\mathbb{C})/GL(n,\mathbb{R})$ mentioned as the homotopy fiber of this functor over $\mathbb{C}^n$, that's what it must be!

John Baez (May 01 2023 at 17:12):

If you want to describe $GL(n,\mathbb{C})/GL(n,\mathbb{R})$ conceptually without saying the word "homotopy fiber", you can call it "the space of all $n$-dimensional real subspaces $V \subseteq \mathbb{C}^n$ with $V + iV = \mathbb{C}^n$."

John Baez (May 01 2023 at 17:13):

Note that not every $n$-dimensional real subspace $V \subset \mathbb{C}^n$ has $V + iV = \mathbb{C}^n$.

John Baez (May 01 2023 at 17:16):

I've seen this same space described as $U(n)/O(n)$. We get this by putting inner products on all our vector spaces before we start doing the computation, and considering only transformations that preserve these inner products.

John Baez (May 01 2023 at 17:18):

If we think of it as $U(n)/O(n)$, we can say it's "the space of all $n$-dimensional real subspaces $V \subseteq \mathbb{C}^n$ on which the imaginary part of the inner product vanishes", since that condition is equivalent to $V + i V = \mathbb{C}^n$, at least for $n$-dimensional real subspaces of $\mathbb{C}^n$.

John Baez (May 01 2023 at 17:19):

And in this guise, this space $U(n)/O(n) \cong GL(n,\mathbb{C})/GL(n,\mathbb{R})$ is called the Lagrangian Grassmannian. (I'm linking to Wikipedia since I helped write that article.)

John Baez (May 04 2023 at 23:48):

Anyone who wants to see how the homotopy fiber (or more officially, "essential fiber") can be used in algebra and geometry can check out my talk slides:

• Symmetric spaces and the tenfold way.

The idea is to show that all 10 infinite series of "compact symmetric spaces" (I say what those are, but they're nice spaces like the ones we've just been seeing here) arise from essential fibers of forgetful functors. I try to explain the idea without category theory at first, since I'll be talking to mathematical physicists, but then at the end I bring in essential fibers.

John Baez (May 04 2023 at 23:48):

Lemma 1 and Lemma 2, near the end, are lifted from our conversation here!

John Baez (May 04 2023 at 23:55):

My talk will eventually be on YouTube and it should be a lot more fun than just these slides... but I may not have time to get into the category theory at the end.