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Stream: deprecated: mathematics

Topic: alternating maps

Oscar Cunningham (Apr 27 2021 at 17:45):

The property 'alternating' is defined on functions out of a product $V\times\dots\times V$. If we pick a basis of $\mathbb{R}^n$ then we get a bijection between $\mathrm{Hom}(\mathbb{R}^n, V)$ and $V\times\dots\times V$ (where there are $n$ copies of $V$). So by translating along this bijection we have a definition of what it means for a function out of $\mathrm{Hom}(\mathbb{R}^n, V)$ to be alternating.

The interesting thing about this idea is that it doesn't depend on the basis we chose for $\mathbb{R}^n$. It's not hard to prove that if a map $a:\mathrm{Hom}(\mathbb{R}^n, V)\to W$ is alternating for one choice of basis then it's alternating for every choice of basis. This makes me think that this is a natural way to think about alternating maps.

In fact given $a:\mathrm{Hom}(\mathbb{R}^n, V)\to W$ one can show that the following are equivalent:

1. For some basis, the corresponding map $V\times\dots\times V\to W$ is alternating.
2. For every basis, the corresponding map $V\times\dots\times V\to W$ is alternating.
3. For every basis, the corresponding map $V\times\dots\times V\to W$ is multilinear.

The last of these points shows that the notion of alternating is quite simple from this point of view. It holds because if $e_i$ and $e_j$ are two basis elements, then one still has a basis if one replaces $e_j$ by $e_j-e_i$. So we get the alternating property from multilinearity for free.

Other facts about alternating maps also become nice from this perspective. For example, for the usual definition we can say that if $a(v_0,\dots,v_{n-1})$ is nonzero then $v_0,\dots,v_{n-1}$ must be linearly independent. From the new point of view this is equivalent to saying that if $a(f)$ is nonzero then $f$ must be injective.

What I'd like to ask is if anyone can see a completely basis independent way of defining 'alternating' from this point of view? I.e. to define what it means to be an alternating map $a:\mathrm{Hom}(U, V)\to W$, without picking a basis of $U$.

One neat application of this would be to apply it when $U$ was a non-free module over a ring. Of course one would still need a restriction on $U$ corresponding to the fact that $n$ must be finite. Presumably $U$ would have to be dualizable, a.k.a. finitely-generated projective. One could then use this to give a nice definition of 'determinant' on dualizable $U$, which currently requires the rather ugly method of arbitrarily choosing a $U'$ such that $U\oplus U'$ is free.

Morgan Rogers (he/him) (Apr 27 2021 at 18:35):

What definition of alternating are you using here? Certainly alternating should be more specific than multilinear!

Oscar Cunningham (Apr 27 2021 at 18:40):

Define the bijection $b:\mathrm{Hom}(\mathbb{R}^n, V)\to V\times\dots\times V$ by $f\mapsto (f(e_0),\dots,f(e_{n-1}))$ where $e$ is the standard basis. Then define $a:\mathrm{Hom}(\mathbb{R}^n, V)\to W$ to be alternating if and only if $b^{-1}\circ a$ is alternating.

The meaning of (3) is that if we have that $a$ such that $b^{-1}\circ a$ is multilinear, and would also be multilinear no matter which basis we used to define $b$, then $a$ is alternating.

Morgan Rogers (he/him) (Apr 27 2021 at 19:12):

Ah, I see the detail I had missed.

John Baez (Apr 27 2021 at 19:45):

If some definition doesn't depend on what basis you pick for $\mathbb{R}^n$, then it also works in a basis-independent way for $U$, where $U$ is any n-dimensional real vector space.

So, it sounds like you're claiming there's a way to define a "multilinear alternating map" from $\mathrm{Hom}(U,V)$ to $W$.

John Baez (Apr 27 2021 at 19:46):

Is there a slick way to do this that never brings in a basis? (You bring in the basis and then claim the choice of basis didn't matter.)

John Baez (Apr 27 2021 at 19:47):

Also, are you claiming we need "alternating" here? That is: can we define a "multilinear" map from $\mathrm{Hom}(U,V)$ to $W$, or only a "multilinear alternating" map?

Oscar Cunningham (Apr 27 2021 at 20:10):

You can define a multilinear map, but only in a basis dependent way. It's 'alternating' that makes it basis independent.

Oscar Cunningham (Apr 27 2021 at 20:11):

John Baez said:

Is there a slick way to do this that never brings in a basis? (You bring in the basis and then claim the choice of basis didn't matter.)

I don't know a way to do it without bringing in a basis. I was hoping someone else would be able to see how to do that!

Morgan Rogers (he/him) (Apr 28 2021 at 06:57):

John Baez said:

Is there a slick way to do this that never brings in a basis? (You bring in the basis and then claim the choice of basis didn't matter.)

You can phrase alternating as "$v_1,\dotsc,v_n$ linearly dependent implies $\alpha(v_1,\dotsc,v_n) = 0$", which is a basis-independent statement. But it still requires you to know what a multilinear map is, so it doesn't really answer the question, sorry :upside_down:

David Wärn (Jul 22 2021 at 12:08):

I'm a bit late to the discussion, but I think you can say that a function $a : \mathrm{Hom}(U, V) \to W$ is alternating if for any $t \in U^\star$, $f \in \mathrm{Hom}(U, V)$, the function $v \mapsto a(f + vt)$ is affine (i.e. some constant plus a linear function in $v$) and moreover $a(f) = 0$ for $f$ non-injective. Basically this is just a restatement of what it means for a function to be multilinear (picking $t \in U^\star$ is like picking one of $n$ arguments).

Gurkenglas (Feb 01 2023 at 22:31):

Resurrecting this for my own insight into alternating maps. Ordered n-tuples are maps from n. Each has a histogram. Unordered n-tuples are subsets of size n. Each has a characteristic function. The pullback is the injections from n. Symmetric maps are those which factor through the histogram. Alternating maps are those which "factor through the injections".
Requesting a note that this is old hat or has a nicer phrasing.