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Is there a notation for the symmetric / antisymmetric subspaces of a tensor power that distinguishes them from the symmetric / exterior power?

Let $V$ be a finite-dimensional vector space over a field $k$, say of characteristic $0$. The symmetric group $S_n$ acts on the tensor power $V^{\otimes n}$ in the obvious way, and this action defines two subspaces of $V^{\otimes n}$, the subspace on which $S_n$ acts via the trivial character and the subspace on which $S_n$ acts via the antisymmetric character.

Question 0: Is the construction of these subspaces functorial in $V$? If it is, are the corresponding functors naturally isomorphic to the symmetric and exterior powers, and if that's true, are the corresponding natural isomorphisms unique?

If the answers to Question 0 turn out more or less like I suspect, we should not regard these subspaces as completely synonymous with the symmetric power $S^n V$ and the exterior power $\Lambda^n V$, respectively, since these are naturally thought of as quotients of $V^{\otimes n}$. (This issue recently came up in another MO question.)

Question 1: Is there an established notation in the literature which respects this distinction?

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$\left(\otimes V\right)_{\mathrm{symm}}$ and $\left(\otimes V\right)_{\mathrm{anti}}$ are what I have encountered. Of course, they don't look particularly slick. – darij grinberg Feb 7 2011 at 21:27
And of course they are functorial. Just check that the image of $\left(\otimes^n V\right)_{\mathrm{symm}}$ under the $n$-th tensor power of a linear map $f:V\to W$ is included in $\left(\otimes^n W\right)_{\mathrm{symm}}$. (That's because the action of $S_n$ is functorial in $V$.) – darij grinberg Feb 7 2011 at 21:28
@Qiaochu: I believe that there are problems in characteristic p (although I could be wrong). I'm also pretty sure that the splitting is not unique, in any case. – Harry Gindi Feb 7 2011 at 21:29
I am pretty sure they are not isomorphic to the symmetric power / the exterior power if the characteristic of $k$ is bad. (Several counterexamples were posted here.) – darij grinberg Feb 7 2011 at 21:30
@darij You shouldn't be answering the question in the comments. It makes it difficult to clear the question from MathOverflow without looking like you're stealing credit from the actual answerer. – Greg Kuperberg Feb 7 2011 at 21:32

This only answers part of your question 0 unfortunatly. The construction is certainly functorial, but the two notions of symmetric/alternating power do not always agree. Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).

If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.

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 I like your choice of notation; I think it is worth adopting. – Qiaochu Yuan Feb 7 2011 at 22:22

1.

For every $k$-vector space $V$, and every $n\in\mathbb N$, the symmetric group $S_n$ acts on the tensor power $V^{\otimes n}$ by permuting the tensorands:

$\sigma\left(v_1\otimes v_2\otimes ...\otimes v_n\right) = v_{\sigma^{-1}\left(1\right)}\otimes v_{\sigma^{-1}\left(2\right)}\otimes ...\otimes v_{\sigma^{-1}\left(n\right)}$ for every $\sigma \in S_n$ and $v_1,v_2,...,v_n\in V$.

Let $V^{\otimes n}_{\mathrm{symm}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts trivially. Let $V^{\otimes n}_{\mathrm{alt}}$ denote the subspace of the tensor power $V^{\otimes n}$ consisting of the elements on which $S_n$ acts by the sign representation.

For every $k$-linear map $f:V\to W$ between two vector spaces, its $n$-th tensor power $f^{\otimes n}:V^{\otimes n}\to W^{\otimes n}$ restricts to a $k$-linear map $f^{\otimes n}_{\mathrm{symm}}:V^{\otimes n}_{\mathrm{symm}}\to W^{\otimes n}_{\mathrm{symm}}$ and a $k$-linear map $f^{\otimes n}_{\mathrm{alt}}:V^{\otimes n}_{\mathrm{alt}}\to W^{\otimes n}_{\mathrm{alt}}$, because the map $f$ commutes with the action of $S_n$ (while $f$ transforms the tensorands, the action of $S_n$ permutes the tensorands). Thus, $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ and $V\mapsto V^{\otimes n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$ are functors.

2.

If $k$ has characteristic $0$, then $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ is isomorphic to $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ as functors, and $V\mapsto V^{\otimes n}_{\mathrm{alt}};\ f\mapsto f^{\otimes n}_{\mathrm{alt}}$ is isomorphic to $V\mapsto \wedge^n\left(V\right);\ f\mapsto \wedge^n\left(f\right)$ as functors. The isomorphisms are given, e. g., in Crawley-Boevey, Lectures on representation theory and invariant theory, ยง6, Lemma 1 and 3.

Of course, multiplying such an isomorphism by a scalar $\neq 0$ yields another isomorphism. This is all the freedom we have: any two isomorphisms between the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ and the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ are equal up to scalar, and similarly for the other pair of functors.

To prove this, we let $P$ be an isomorphism from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$. Let $\lambda\in k$ be defined by $P_k\left(1\otimes 1\otimes ...\otimes 1\right)=\lambda 1\cdot 1\cdot ...\cdot 1$, where $P_k$ is the isomorphism $P$ at the object $V=k$, and $\cdot$ denotes the multiplication in the symmetric algebra (because it is commutative).

Since $k$ has characteristic $0$, the space $V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes\dots\otimes v$ for $v\in V$. (This is Lemma 7 from Crawley-Boevey's above-mentioned text, sent back to $V^{\otimes n}_{\mathrm{symm}}$ from $\mathrm{S}^n\left(V\right)$.) We are now going to prove that $P_V\left(v\otimes v\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot \ldots\cdot v$ for every $v\in V$.`

In order to show this, let $f:k\to V$ be a vector space homomorphism given by $f\left(1\right)=v$. The functoriality of $P$ now yields

$P_V\left(f^{\otimes n}_{\mathrm{symm}}\left(1\otimes 1\otimes\ldots\otimes 1\right)\right)=\left(\mathrm{S}^n\left(f\right)\right)\left(P_k\left(1\cdot 1\cdot\ldots\cdot 1\right)\right)$.

This rewrites as $P_V\left(v\otimes v\otimes \ldots\otimes v\right)=\lambda v\cdot v\cdot \ldots\cdot v$, and we are done.

This yields (since the space $V^{\otimes n}_{\mathrm{symm}}$ is generated by the tensors $v\otimes v\otimes\ldots\otimes v$ for $v\in V$) that the map $P_V$ is just the canonical projection from $V^{\otimes n}_{\mathrm{symm}}$ to $\mathrm{S}^n\left(V\right)$, multiplied with the scalar $\lambda$. Since $\lambda$ does not depend on $V$, this shows us that our isomorphism $P$ from the functor $V\mapsto V^{\otimes n}_{\mathrm{symm}};\ f\mapsto f^{\otimes n}_{\mathrm{symm}}$ to the functor $V\mapsto \mathrm{S}^n\left(V\right);\ f\mapsto \mathrm{S}^n\left(f\right)$ is the projection isomorphism times $\lambda$. In other words, all the freedom we have to choose this isomorphism is the freedom of choosing the scalar factor to multiply with. The same argument works for the other pair of functors.

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Oh, and if you are wondering why the whole text is in a "code" section: I can't be bothered to put every single formula that contains indices in backticks, and there doesnt seem to be a good way to figure out whether this is necessary just by looking at the TeX code. We need an auto-backtick-tex bot... – darij grinberg Feb 7 2011 at 22:37
As Qiaochu's MathOverflow source shows, there is no problem, or no problem any longer, with underscores. You only need the backticks at the most for asterisks, and Qiaochu's solution is to use \ast. – Greg Kuperberg Feb 7 2011 at 22:49
Not fixed. See my revision 2. – darij grinberg Feb 7 2011 at 22:52
I, on the other hand, can't handle reading in the giant "code" font, so I have taken the liberty to add the back-ticks where needed. I didn't back-tick everything, so I hope I didn't miss a \{. – Theo Johnson-Freyd Feb 8 2011 at 3:25
@Greg: darij's problem is less with * than with _. – Theo Johnson-Freyd Feb 8 2011 at 3:26

The subspace of $\Sigma_n$-invariants of $V^{\otimes n}$ is called the $n$th divided power of $V$ (at least when $V$ is a free module).

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Okay, so I think this is what's going on. The elements of $V^{\otimes n}$ are naturally identified with multilinear functions on $V^{\ast}$. Among these functions are the alternating multilinear functions on $V^{\ast}$, which are naturally identified with the elements of $\Lambda^n(V^{\ast})^{\ast}$, so this gives a natural inclusion $\Lambda^n(V^{\ast})^{\ast} \to V^{\otimes n}$. The situation is similar for symmetric tensors. I think this inclusion is what I'm looking for.

One can compose the above inclusion with the quotient $V^{\otimes n} \to \Lambda^n(V)$, and this gives a natural map $\Lambda^n(V^{\ast})^{\ast} \to \Lambda^n(V)$. In characteristic zero this map is an isomorphism, but in positive characteristic there are problems, and in any case it behaves in a slightly unexpected way with respect to a basis of $V$ (because of the issues Greg Kuperberg brought up in the linked MO question).

Here is what happens in the simplest nontrivial case. Let $V$ be two-dimensional with basis $e_1, e_2$. Then $V^{\otimes 2}$ inherits a natural basis $e_i \otimes e_j, 1 \le i, j \le 2$ and the image of this basis gives a basis $e_1 \wedge e_2$ of $\Lambda^2 V$. The dual $V^{\ast}$ inherits a dual basis $e_1^{\ast}, e_2^{\ast}$ giving a basis $e_1^{\ast} \wedge e_2^{\ast}$ of $\Lambda^2 V$, and dualizing one more time gives a dual basis $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$. The natural inclusion above sends $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$ to $e_1 \otimes e_2 - e_2 \otimes e_1$.

But now the natural map $\Lambda^2(V^{\ast})^{\ast} \to \Lambda^2(V)$ sends $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$ to $2 e_1 \wedge e_2$. So natural bases do not behave in the expected way with respect to these constructions, and one must insert some factorials...

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I would advise everyone to steer clear of dual spaces unless necessary. You can do all the useful stuff with double-dual spaces if $V$ is a finite-dimensional vector space over a field, but the functorial isomorphisms $V^{\otimes n}_{\mathrm{symm}}\to \mathrm{Sym}^n V$ and $V^{\otimes n}_{\mathrm{alt}}\to \wedge^n V$ are well-defined isomorphisms for any $k$-module $V$, where $k$ is a $\mathbb Q$-algebra. No finite-dimensionality or $k$ being a field is required. It is just the canonical projection in one direction, and an average rsp. alternating average over the symmetric group in the other. – darij grinberg Feb 7 2011 at 23:27
@darij: fair point. – Qiaochu Yuan Feb 7 2011 at 23:41