# Signs in the natural map $\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$

Let $V$ be a finite-dimensional vector space over a field $\Bbbk$. Let $V^*$ denote its dual. I strongly suspect that there is a natural map $$\Lambda^k V \otimes \Lambda^k V^* \to \Bbbk$$ that looks something like $$v_1 \wedge \dotsb \wedge v_k \otimes \alpha_1 \wedge \dotsb \wedge \alpha_k \mapsto \sum_{\sigma} {\operatorname{sgn} \, \sigma}\prod_i \alpha_i(v_{\sigma(i)}).$$ What does the correct, natural formula look like? In particular, what is the correct sign convention?

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yes, that's the right formula (your sign is correct) –  user8268 Jun 8 '11 at 20:15
I think you mean just $\operatorname{sgn} \sigma$, not $(-1)^{\operatorname{sgn} \sigma}$. –  Raeder Jun 8 '11 at 21:33
Raeder: You are, of course, correct. I've fixed it. –  Charles Staats Jun 8 '11 at 21:48
It is just the determinant of the matrix $(\alpha_i(v_k))_{i,k}$. –  Martin Brandenburg Feb 3 at 14:02

While at the vector space level, the pairing might seem slightly forced, we can derive it naturally by adding structure.

Given a vector space $V$, we have a graded commutative ring $\bigwedge V = \bigoplus_i \bigwedge^i V$.

Given $\phi\in V^*$, it naturally extends to a (graded) derivation $d_{\phi}$ of degree $-1$ on $\bigwedge V$. Since $d_{\phi}^2=0$ and $d_{\phi+\psi}=d_{\phi}+d_{\psi}$, we can extend the action of $V^*$ to an action of $\bigwedge V^*$. The pairing is just the action restricted to a single degree.

Elaboration on the constructions:

First, we need to see that specifying a derivation by how it acts on generators is actually well defined. Note that, $\bigwedge V = T(V)/(v\otimes v\mid v\in V)$ is a quotient of the tensor algebra, Given any $\phi \in V^*$, we can define a derivation $d_{\phi}$ of $T(V)$ extending $\phi$, and because every element of $T(V)$ can be written in a unique way, such a derivation is well defined. For any degree $-1$ derivation $d$ we have $d(v^2)=(dv)v-v(dv)=0$, and so $d$ vanishes on the ideal defining $\bigwedge V$, and thus passes to a well defined map there.

To see that derivations extend to an action of $\bigwedge V^*$, we have that if $d:V\mapsto A$ is a linear map of a vector space into an algebra such that $d^2(v)=0$ for every $v\in V$, then there exists a unique map $\bigwedge V \to A$ extending $d$. However, care must be taken here, as we want $A$ to be a graded algebra and we want $d(V)\subset A_1$.

Unfortunately, because we wish our map to take values in $\operatorname{End}_k(\bigwedge V)$, which is not commutative, we can't just use the universal property of $\bigwedge V$ being the free graded commutative algebra generated in degree $1$, and we have to* do things at the level of the tensor algebra and show that things descend.

All these are related to various structures present in differential forms and vector fields, and the interaction between them (e.g. Lie derivatives), which can be extended further to structures in Hochschild homology and cohomology. There are also analogies to be made between cup and cap products in algebraic topology.

Other related ideas worth looking into are the variants of the Schouton bracket.

Note that most of the related structures are not entirely linear, and that the structure we have here here is merely a linear approximation to them.

*No, we probably don't have to. I just can't think of a cleaner way to do it at the moment. If anybody has suggestions, please let me know.

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Oh, I see now. This is quite nice. –  Qiaochu Yuan Jun 8 '11 at 22:28

I can construct this map abstractly, but I want to convince you that it isn't completely natural. Let's work in more generality: suppose $A \otimes B \to \mathbb{k}$ is a bilinear pairing. If I want to replace $A$ with some quotient $A/A'$, what's the natural thing to do to the pairing? If $A, B$ are finite-dimensional, then giving a bilinear pairing is the same as giving a map $A \to B^{\ast}$. If I want to replace $A$ with a quotient, then the natural thing to do is to send this map to the induced map $A/A' \to B^{\ast}/\text{im}(A')$. But dualizing the quotient map $B^{\ast} \to B^{\ast}/\text{im}(A')$ gives an inclusion

$$\left( B^{\ast}/\text{im}(A') \right)^{\ast} \to B.$$

The LHS is the subspace of $B$ annihilated by every element of $A'$. So contrary to intuition, the natural thing to do is not to replace $B$ by a quotient. Note that this recipe has the desirable property that if the old pairing is nondegenerate, so is the new pairing.

Now let $A = V^{\otimes k}, B = (V^{\ast})^{\otimes k}$. These spaces are equipped with a canonical pairing $A \otimes B \to k$. If I want to replace $A$ by its quotient $\Lambda^k(V)$, then the above recipe tells me that the correct thing to do is to replace $B$ by a subspace, which turns out to be precisely the subspace of antisymmetric tensors $\text{Alt}^k(V^{\ast}) \subset (V^{\ast})^{\otimes k}$. Note that this is not abstractly the same thing as $\Lambda^k(V^{\ast})$. So the correct replacement pairing is

$$\Lambda^k(V) \otimes \text{Alt}^k(V^{\ast}) \to \mathbb{k}$$

which I believe is nondegenerate in characteristic greater than $2$. In addition, there is a natural map

$$\text{Alt}^k(V^{\ast}) \to (V^{\ast})^{\otimes k} \to \Lambda^k(V^{\ast})$$

which I believe is an isomorphism in characteristic greater than $k$ but is zero in characteristic less than or equal to $k$. The problem is that the space on the left is spanned by elements of the form

$$\sum_{\pi \in S_k} \text{sgn}(\pi) e_{\pi(1)} \otimes e_{\pi(2)} \otimes ... \otimes e_{\pi(k)}$$

where $e_1, ... e_k$ are a $k$-element subset of a basis of $V^{\ast}$, and the image of this element in $\Lambda^k(V^{\ast})$ is $k! e_1 \vee e_2 \vee ... \vee e_k$ which vanishes if $k! = 0$.

Punchline: if you use only the natural maps above, I think the pairing you want is only natural in characteristic greater than $k$ and it's given by $\frac{1}{k!}$ times what you wrote. As far as sign convention, this is all a matter of what you think the natural pairing

$$V^{\otimes k} \otimes (V^{\ast})^{\otimes k} \to \mathbb{k}$$

is. Do you think it's given by evaluating the middle two factors on each other, then the next middle two, and so forth, or do you think it's given by evaluating the first factor in $V^{\otimes k}$ on the first factor in $(V^{\ast})^{\otimes k}$, and so forth? You use the second convention in your post but to me the first convention is more natural (at least it generalizes in a less annoying way to a symmetric monoidal category with duals).

The above discussion is closely related to another confusing property of the exterior power, which is that if $V$ has an inner product then the natural space which inherits an inner product from $V$ is not $\Lambda^k(V)$ but $\text{Alt}^k(V)$, and people don't always use the canonical map between these spaces; for example people sometimes want the exterior product of orthogonal unit vectors to be a unit vector, but that is actually false if you only use natural maps, and it's necessary to normalize a map somewhere (either the identification above or, equivalently, the antisymmetrization map).

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Thanks, this is a nice way to look at it! I assume, based on which factors are paired with which, that the formula I gave corresponds to the second of the two conventions you describe? –  Charles Staats Jun 8 '11 at 21:17
@Charles: yes, I think so. –  Qiaochu Yuan Jun 8 '11 at 21:19
I'm not sure I agree that your recipe from the first paragraph is indeed the most natural thing. I haven't thought enough about this stuff, but Brian Conrad wrote extensive notes: math.stanford.edu/~conrad/diffgeomPage/handouts/tensor.pdf Here he seems to argue that the map from Charles is indeed natural. –  wildildildlife Jun 8 '11 at 21:46
@wildildildlife: note that he did not write down the symmetric or exterior pairing in an invariant way: his definition requires that one first defines the induced pairing on pure tensors whereas mine doesn't. I don't even know if this can be done. His comment in the very last paragraph is spot on regarding the symmetric and exterior products, but I'm going to have to disagree with him about induced bilinear pairings until he can construct his maps without defining them on pure tensors first. –  Qiaochu Yuan Jun 8 '11 at 22:00
Qiaochu: The purity of a tensor is, in fact, an invariant property. The set of pure tensors form a Zariski-closed subset X of the space of all tensors. In fact, X is the cone of the Grassmanian under the Plucker embedding. –  Charles Staats Jun 8 '11 at 22:08

As I remarked in one of the comments, a nice write-up about tensor and exterior pairings is this one by Brian Conrad (as part of a series of handouts to be found at his website).

His approach is to let a general bilinear pairing $B:V\times W\to k$ yield a pairing $V^{\otimes n}\times W^{\otimes n}\to k$ given by $(\otimes v_i,\otimes l_j)\mapsto \prod_{i=1}^n B(v_i,l_i)$. Under the natural conditions (invariance under swaps, or vanishing if a sequence of inputs `stammers'), this induces parings on the symmetric, or exterior algebras.

In particular, applied to the evaluation pairing $B:V\times V^*\to k$ this induces the desired pairing $\bigwedge^n(V)\times \bigwedge(V^*)\to k$ given by $(\wedge v_i,\wedge l_j)\mapsto \det(l_i(v_j))$.

(This is a very short summary; his explanation is much better and extensive so read it yourself :))

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I can only make sense of this construction by writing it in greater generality: we need to let $B : V \times W \to \mathbb{k}$ be a pairing which is invariant under the action of a finite group $G$ on $V, W$ and then I think it ought to be true that this induces a bilinear form on isotypic quotients, but I haven't worked out the details. –  Qiaochu Yuan Jun 9 '11 at 1:21
wildildildlife: I've accepted Aaron's answer because I think he put the most work into it, but I would again like to say that I found this reference very helpful. –  Charles Staats Jun 11 '11 at 20:06
@Charles: Sure! Credits for this answer should go to Brian Conrad anyway :) –  wildildildlife Jun 11 '11 at 20:22
@Qiaochu: Yes, essentially this has nothing to do with vector spaces or alike, it works in any cocomplete tensor cateogory. –  Martin Brandenburg Feb 2 at 14:11