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Stream: deprecated: id my structure

Topic: Inner product on an algebra with nice multicategorical prop?

Kevin Arlin (Sep 05 2023 at 18:09):

I'm wondering about the situation in a multicategory where we have a magma $A$ with multiplication $\mu:(A,A)\to A$ together with a binary map $\nu:(A,A)\to X,$ and we ask about the existence of a factorization $\ell:A\to X$ of $\nu$ through $\mu.$ If $A$ is unital, then such an $\ell$ must be unique if it exists.

Kevin Arlin (Sep 05 2023 at 18:10):

I thought to consider what this looks like in the multicategory of vector spaces over some field $k.$ Then $A$ is a $k$-algebra, not necessarily associative or unital, and if we set $X=k$ then we're asking about the existence of a bilinear form on $A$ which factors through $A$'s multiplication, so that $\langle a_1,a_2\rangle= \ell(a_1a_2)$ for some linear form $\ell.$

Kevin Arlin (Sep 05 2023 at 18:10):

This seems like a natural enough condition, but I'm not able to find occurrences in so-called real life. In general, assuming the multiplication is surjective, then $\ell$ must be defined by $\ell(a_1a_2)=\langle a_1,a_2\rangle,$ so that the condition is equivalent to the inner product being constant on fibers of the product. In particular if the algebra is unital then we can define $\ell(a)=\ell(1\cdot a)=\langle 1,a\rangle$ and if it's associative then we're asking for an $A$-billinear inner product, since we have $\langle a,bc\rangle=L(abc)=\langle ab,c\rangle.$ I'm currently confused about what happens in the case of $\mathbf C.$

Kevin Arlin (Sep 05 2023 at 18:10):

Moving away from unital and associative algebras, I thought the idea of invariant inner product on a Lie algebra might produce an example, but in the Lie case my condition reduces instead to $\langle a_1,[a_2,a_3]\rangle=\langle a_1a_2,a_3\rangle-\langle a_2,a_1a_3\rangle$, i.e. it's just the condition that the inner product satisfies the Jacobi identity. And that's not what people mean when they define invariant inner products.

Kevin Arlin (Sep 05 2023 at 18:11):

Can anybody think of familiar examples of this behavior occurring, whether in $k$-algebras or in some other practically familiar multicategory? Alternatively, is there some reason why this is a weird structure to think about?

Kevin Arlin (Sep 05 2023 at 18:37):

Well, I guess all that happens in the $\mathbf C$ case, tautologically, is that you can factor a bilinear form through the multiplication if and only if it's of the form $\ell(xy)$ for some linear form $\ell:\mathbf C\to \mathbf R.$ Thus there's a 2-dimensional space of bilinear forms on $\mathbf C$ that work this way, out of four total dimensions of such forms. And in general, if $A$ is an $n$-dimensional algebra with a surjective multiplication, then there's an $n$-dimensional space of bilinear forms that can be made to respect that multiplication out of the $n^2$-dimensional space of bilinear forms all in all.

But that still doesn't help me with the question of whether there are important examples of bilinear forms like this!

John Baez (Sep 05 2023 at 19:52):

Kevin Arlin said:

I'm wondering about the situation in a multicategory where we have a magma $A$ with multiplication $\mu:(A,A)\to A$ together with a binary map $\nu:(A,A)\to X,$ and we ask about the existence of a factorization $\ell:A\to X$ of $\nu$ through $\mu.$ If $A$ is unital, then such an $\ell$ must be unique if it exists.

This seems like a natural enough condition, but I'm not able to find occurrences in so-called real life.

This happens whenever you have a semisimple associative unital algebra over a field $k$ - a situation of great importance to old-school algebraists, in part because the algebra of $n \times n$ matrice is an example, and in part because they're completely classified by the Wedderburn-Artin theorem.

I won't always "associative unital" since that's the only case I'm really interested in now, though I will at times digress.

John Baez (Sep 05 2023 at 19:54):

First a few more examples. Over $k = \mathbb{C}$ the only semisimple algebras are finite direct sums of the $n \times n$ matrix algebras $M(n, \mathbb{C})$.

John Baez (Sep 05 2023 at 19:55):

Over $k = \mathbb{R}$ the only semisimple algebra are finite direct sums of $M_n(\mathbb{R}), M_n(\mathbb{C})$ and $M_n(\mathbb{H})$, where $\mathbb{H}$ is the quaternions.

John Baez (Sep 05 2023 at 19:56):

The Wedderburn-Artin theorem explains and generalizes all this.

Kevin Arlin (Sep 05 2023 at 19:57):

Thanks! Is the form you're thinking of on matrix algebras $\mathrm{tr}(A^*B)$ or something like that?

John Baez (Sep 05 2023 at 19:58):

Wait and see. For any finite-dimensonal algebra (even not necessarily associative or unital!) $A$ over a field $k$, there's a god-given bilinear form $b: A \otimes A \to k$ given by

$b(x \otimes y) = \mathrm{tr}(L_x L_y)$

John Baez (Sep 05 2023 at 19:59):

Here $L_x: A \to A$ is left multiplication by $x \in A$, and $\mathrm{tr}$ is the usual trace of linear maps from the vector space $A$ to itself.

John Baez (Sep 05 2023 at 20:00):

$b$ is called the Killing form, and it's symmetric:

$b(x \otimes y) = b(y \otimes x)$

John Baez (Sep 05 2023 at 20:01):

The Killing form always gives a linear map $\ell : A \to A^\ast$ by

$\ell(x) (y) = b(x \otimes y)$

John Baez (Sep 05 2023 at 20:03):

We say an algebra is semisimple if $\ell$ is an isomorphism. This term is commonly used when $A$ is an associative unital algebra - our case of main interest here. It's also widely used for Lie algebras. But we could make this definition for any algebra, not necessarily associative or unital.

John Baez (Sep 05 2023 at 20:06):

Semisimple finite-dimensional algebras that are associative and unital are classified by Wedderburn-Artin, and for these the Killing form $b : A \otimes A \to k$ does what you want: it factors through a linear map $\epsilon : A \to k$.

John Baez (Sep 05 2023 at 20:08):

What is this map? I'll tell you! The linear map

$\iota : k \to A$

sending $1 \in k$ to the multiplicative unit $1 \in A$ has as its adjoint a linear map

$\iota^\ast : A^\ast \to k^\ast$

John Baez (Sep 05 2023 at 20:09):

But $k^\ast$ is isomorphic to $k$ in a standard way, and since we're assuming $A$ is semisimple we have an isomorphism $\ell : A \to A^\ast$.

John Baez (Sep 05 2023 at 20:09):

Thus $\iota^\ast$ gives a linear map

$\epsilon : A \to k$

And this is what we want! It's often called the counit of our semisimple algebra.

John Baez (Sep 05 2023 at 20:13):

Kevin Arlin said:

Thanks! Is the form you're thinking of on matrix algebras $\mathrm{tr}(A^*B)$ or something like that?

No, it's $\mathrm{tr}(L_a L_b)$, where $L_a : A \to A$ is defined as above. To define this we don't need to know that our algebra is a matrix algebra, just a finite-dimensional algebra. We don't even need to know our algebra is associative or unital!

In the case when our algebra is a matrix algebra $\mathrm{tr}(L_a L_b)$ proportional but not equal to what you'd call $\mathrm{tr}(A B)$. But note: there's definitely no $A^\ast$ going on here.

James Deikun (Sep 05 2023 at 20:25):

What happens if you use $\mathrm{tr}(R_{a}R_b)$?

John Baez (Sep 05 2023 at 20:27):

You get the same thing. Checking this is a good way to see what's really going on here.

Kevin Arlin (Sep 05 2023 at 20:27):

Fantastic, thanks again! OK, and to trace it through, let's see... for a semisimple algebra you get $\epsilon$ from $A\to A^*\to k^*\to k$ sending $x\mapsto (y\mapsto \mathrm{Tr}((LxLy))\mapsto (t\mapsto \mathrm{Tr}(LxL\iota(t))\mapsto \mathrm{Tr}(LxL\iota(1))=\mathrm{Tr}(Lx).$ As for the proportionality, I guess that if $x$ is a matrix then $\mathrm{Tr}(Lx)=n\mathrm{Tr}(x)$.

John Baez (Sep 05 2023 at 20:29):

Yes, if $x$ is an $n \times n$ matrix then $\mathrm{tr}(L_x) = n \mathrm{tr}(x)$

John Baez (Sep 05 2023 at 20:30):

The idea here is that matrices happen to act as linear transformations of vectors, but every algebra acts as linear transformations of itself.

James Deikun (Sep 05 2023 at 20:37):

If I'm right the Killing form of $\mathbb{H}$ over $\mathbb{R}$ should be proportional to $\mathrm{Re}(xy)$...

James Deikun (Sep 05 2023 at 20:39):

And the same for $\mathbb{O}$ over $\mathbb{R}$ ...

John Baez (Sep 06 2023 at 07:31):

Yes, those are correct. Note that these are not positive definite, so they're not inner products, but they are nondegenerate symmetric bilinear forms. The Killing form is always symmetric (fun to check using string diagrams), and its nondegenerate for semisimple algebras (by definition). But the question you raise is: when does the Killing form factor through a map $A \to k$? And I think it always does when your algebra has a unit.

John Baez (Sep 06 2023 at 07:33):

In the middle of the night I woke up and tried to prove $\mathrm{tr}(L_x L_y) = \mathrm{tr}(R_x R_y)$ and had trouble, though it's pretty easy for commutative algebras, for associative algebras and for Lie algebras.

John Baez (Sep 06 2023 at 07:34):

Then I broke down and used structure constants.

John Baez (Sep 06 2023 at 07:35):

In terms of those the usual "left" Killing form is

$b_{i j} = c_{i k}^\ell c_{j \ell}^k$

John Baez (Sep 06 2023 at 07:36):

The "right" Killing form you invented is

$b'_{ij} = c_{k i}^\ell c^{k}_{\ell j}$

John Baez (Sep 06 2023 at 07:37):

I'm using traditional tensor notation here, and the [[Einstein summation convention]], which are the way people drew string diagrams before it was easy to draw string diagrams in papers.

John Baez (Sep 06 2023 at 07:40):

I don't see any reason why these should be equal except in the 3 special cases I mentioned. For commutative algebras we have symmetry:

$c_{ik}^\ell = c_{ki}^\ell$

while for Lie algebras we have antisymmetry:

$c_{ik}^\ell = -c_{ki}^\ell$

Either of these make it easy to show $b_{ij} = b'_{ij}$.

John Baez (Sep 06 2023 at 07:47):

Associativity is equivalent to this:

$c_{i k}^\ell c_{\ell j}^m = c_{i \ell}^m c^\ell_{k j}$

John Baez (Sep 06 2023 at 08:12):

Hmm, now I'm not even succeeding in showing the left and right Killing forms agree for associative algebras! I've got a page of computations here, but I'm not getting it!

John Baez (Sep 06 2023 at 08:14):

I may be getting tangled up somehow.

But for now, I take back my claim: I'm reduced to saying the left and right Killing forms agree for algebras with a commutative product or with an anticommutative product.