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Stream: learning: questions

Topic: n.t. commutativity square

Jan Pax (Aug 11 2022 at 08:28):

I need the naturality square diagram for the following problem; I cannot carry it out myself. For a real inner product space $( V, \langle\cdot,\cdot\rangle)$ there is a natural map from $V$ to $V^*$. Namely, vector $u \in V$ goes to linear functional $\phi_u \in V^*$ defined by $\phi_u(x) = \langle x,u\rangle$. In case $V$ is finite-dimensional, this is an isomorphism from $V$ to $V^*$. Make it a (contravariant) functor as usual: if $T : V_1 \to V_2$ is a linear transformation, then $T^* : V_2 \to V_1$ satisfies
$\langle T(x),u \rangle = \langle x,T^*(u)\rangle .$
That is, with functionals as above, $T^* : V_2^* \to V_1^*$ is
$T^*(\alpha) = \alpha\circ T,\quad\text{for }\alpha \in V_2^*$. Now, I want to show that ${}^*$ is a natural functor. The square should have vertices $V_1,V_1^*,V_2,V_2^*$. Though I know the vertices, I cannot draw it myself.

Peter Arndt (Aug 11 2022 at 12:09):

What is a "natural functor"?

Spencer Breiner (Aug 11 2022 at 13:01):

You are probably running into problems because you are comparing a covariant functor (id) with a contravariant functor (*). That makes the diagrams look a little different, but if you follow your nose you will get a commuting square with one arrow on one side and three on the other. See Martin Brandenburg's answer here.

Don't forget, the isomorphism is only natural for linear functions that preserve the inner product.

Jan Pax (Aug 11 2022 at 15:36):

I really got stuck: Here, I should have used some functors $F$ and $G$ but I do not know which ones... $\begin{array}{ccc} V_1 & \rightarrow & V_1^* \\ {T} \downarrow && \ \ \ \uparrow {T^*}\\ V_2 & \rightarrow & V_2^* \end{array}$ I just cannot transfer the abstract notion of a natural transformation into my current setting. Perhaps $\mathbf{Id}\to ()^*$ is a natural transformation ??

Mike Shulman (Aug 11 2022 at 17:27):

Did you read the linked answer? It's not a natural transformation in the ordinary sense, but it is a special sort of [[dinatural transformation]].

Jan Pax (Aug 11 2022 at 18:30):

Yes, I've read it. But I just got an answer that it is not a naturality square at all,quoted here: No, it certainly doesn't commute. If $V_1$ is $\mathbb R^2$ with the Euclidean norm, then we can think of $V_1$ as $1 \times 2$ matrices, and the correspondence $^*$ lets us think of $V_1^*$ as $1\times 2$ matrices. The horizontal arrows map each vector to its transpose. If $T : V_1 \to V_2$ is represented by a $2 \times 2$ matrix acting on the left, then $T^*$ is the transpose matrix acting on the right. For example, if $T = 2I$, twice the identity, then in the diagram the arrow across the top is the transpose, while the composition around the other three sides is $4$ times the transpose.

Mike Shulman (Aug 11 2022 at 18:50):

Right, it only commutes when T is orthogonal, i.e. the correct sort of "morphism of inner-product spaces".

Tobias Schmude (Aug 24 2022 at 09:37):

We have to restrict to orthogonal maps to get a "natural" isomorphism between functors of different variance anyway. So in this case we can see it as a natural isomorphism $(-)^{-1} \Rightarrow (-)^*$ in the usual sense, where the left hand side is the contravariant inversion endofunctor on the groupoid of real inner product spaces with orthogonal maps. @Jan Pax: This is quite restricted and pretty tautological, but maybe it's more to your taste?

This holds rather generally: if we have such a "natural" isomorphism between functors of different variance this forces them to land in the groupoid core of their codomain, so such a transformation is equivalently given by a usual natural isomorphism between one functor and the other composed with inversion. So in this case this concept does not yield anything new.

Jan Pax (Aug 24 2022 at 09:45):

Thank you for your reply, Tobias. What is a grupoid of real inner product spaces and a grupoid center in your intended sense ?

Tobias Schmude (Aug 24 2022 at 09:47):

I just mean the restriction of the category of inner product spaces and linear maps to orthogonal maps. They're invertible, so this is a groupoid.

John Baez (Aug 24 2022 at 20:24):

In general "the category of X's" means the category with X's as objects and some supposedly obvious sort of morphisms between them. If someone says this and you don't know what the obvious morphisms are, you ask. Then, "the groupoid of X's" means the category with X's and invertible morphisms between them. Often this is much easier to guess than the category of X's - we've discussed this here not long ago.

For example, it's obvious (to anyone who knows this game) that an invertible morphisms between real inner product spaces is a linear bijection that preserves the inner product.

John Baez (Aug 24 2022 at 20:26):

Anyone who is not used to this game should try to figure out what categories these are:

• the groupoid of sets
• the groupoid of groups
• the groupoid of partially ordered sets

These are phrases that people around here will often say.

Mike Shulman (Aug 24 2022 at 20:30):

John Baez said:

In general "the category of X's" means the category with X's as objects and some supposedly obvious sort of morphisms between them.... Then, "the groupoid of X's" means the category with X's and invertible morphisms between them.

And this groupoid of Xs is sometimes called the core (not center) of the category of Xs, as Tobias did.

John Baez (Aug 24 2022 at 20:45):

Right. I was wondering what that "center" business was. People often talk about the center of a monoidal category (or other things), but that's something else.