Let $V$ be a finite-dimensional vector space over a field $k$, $v_1, ... v_n \in V$ a set of vectors, and $f_1, ... f_n \in V^{\ast}$ a set of covectors. Up to permutation, there seem to be at least two "natural" choices of pairing

$$\Lambda^n(V) \otimes \Lambda^n(V^{\ast}) \to k.$$

One is given by extending

$$(v_1 \wedge ... \wedge v_n) \otimes (f_1 \wedge ... \wedge f_n) \to \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n f_i(v_{\sigma(i)}).$$

It can be found, for example, in these notes by Brian Conrad, and it has the desirable property that if $e_1, ... e_n$ are part of a basis of $V$ and $e_1^{\ast}, ... e_n^{\ast}$ are the corresponding parts of the dual basis, then $e_1 \wedge ... \wedge e_n$ is dual to $e_1^{\ast} \wedge ... \wedge e_n^{\ast}$.

The other is given by extending

$$(v_1 \wedge ... \wedge v_n) \otimes (f_1 \wedge ... \wedge f_n) \to \frac{1}{n!} \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n f_i(v_{\sigma(i)}).$$

It has the disadvantage of not being defined in characteristic $n$ or less, but of the two, this is the one I know how to define functorially; one functorial construction is detailed in my answer to this math.SE question and another is given by starting with the induced pairing $V^{\otimes n} \otimes (V^{\ast})^{\otimes n} \to k$, restricting to antisymmetric tensors $\text{Alt}^n(V) \otimes \text{Alt}^n(V^{\ast}) \to k$, and then inverting the canonical map $\text{Alt}^n(V) \to V^{\otimes n} \to \Lambda^n(V)$ as well as the corresponding map for the dual space (these maps being isomorphisms only in characteristic greater than $n$).

In the linked math.SE answer, Aaron indicates that the first pairing is a restriction of a more general functorial action of $\Lambda(V^{\ast})$ on $\Lambda(V)$. I initially thought this construction worked almost entirely on the basis of certain universal properties, but Aaron ends up having to check that certain conditions are met by hand; either way, this construction isn't ideal as it doesn't seem to treat $V$ and $V^{\ast}$ symmetrically. So:

(How) can I functorially construct the first pairing in a way that treats $V$ and $V^{\ast}$ symmetrically? Is it the case that the only functorial constructions of the first pairing treat $V$ and $V^{\ast}$ asymmetrically?

(By "functorially construct" I mean, at a minimum, that you don't ever have to specify what happens to pure tensors and extend.)

I put "how" in parentheses because I'm not convinced that this is possible. One way to do it is to fix a nice embedding $\Lambda^n(V) \to \text{Alt}^n(V)$. Let's suppose that we send $v_1 \wedge ... \wedge v_n$ to

$$c_n \sum_{\sigma \in S_n} \text{sgn}(\sigma) v_{\sigma(1)} \otimes ... \otimes v_{\sigma(n)}$$

where $c_n$ depends only on $n$ (in particular, it shouldn't depend on $V$). Then, if I'm not mistaken, the induced pairing $\Lambda^n(V) \otimes \Lambda^n(V^{\ast}) \to k$ is given by extending

$$(v_1 \wedge ... \wedge v_n) \otimes (f_1 \wedge ... \wedge f_n) \to c_n^2 n! \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n f_i(v_{\sigma(i)})$$

so the only way to recover the first pairing is if $c_n = \frac{1}{\sqrt{n!}}$, which is terrible since it depends on this particular element existing in $k$. The above computation reminds me of an issue with the normalization of the discrete Fourier transform on $\mathbb{Z}/n\mathbb{Z}$, the problem being that there are at least two natural measures (counting measure and the unique invariant probability measure) which are Fourier dual under the discrete Fourier transform, and the only one which is Fourier self-dual assigns each point the measure $\frac{1}{\sqrt{n}}$.

The discussion at this MO question seems related, although it doesn't seem to immediately answer my question. In the geometric context, I am asking what the natural pairing is between differential forms and polyvector fields. People seem to agree that there is one, but I don't know where to find an authoritative opinion on what it is. In geometric language, I guess want to think of polyvector fields as "infinitesimal cubes" on a manifold, so the correct pairing should be given by "integration."

Can this be made precise, and does it give the first pairing or the second one?

AlgebraChapter III, Section 11 is a convenient reference for the approach that Ekedahl sketches. $\endgroup$ – user2490 Jun 16 '11 at 22:2310more comments