The interior product and the isomorphism $\bigwedge^k(V^*)\otimes\bigwedge^n(V)\cong\bigwedge^{n-k}(V)$

Question

Let $V$ be an $n$-dimensional vector space. According to Wikipedia, there is an isomorphism $\bigwedge^k(V^*)\otimes\bigwedge^n(V)\cong\bigwedge^{n-k}(V)$. The explanation is that for $\alpha \in \bigwedge^k(V^*)$ and $\sigma \in \bigwedge^n(V)$, the isomorphism is given by $i_{\alpha}\sigma$ where $i_{\alpha}$ denotes the interior product (or multiplication) with $\alpha$. I'm confused by this. I've only ever seen the interior product of a form by a $1$ covector. How does one define $i_{\alpha}\sigma$?

Answer 1

Let $\{e_i\}$ denote a basis of the finite dim. vector space $V$. With $\{w_i\}$ we denote the dual basis on $V^{*}$. The contraction $i:\wedge(V^{*})\otimes\wedge V\rightarrow \wedge V$ is defined as follows. (a small note: if you want $i$ to be of degree $0$, you must reverse the grading on the exterior algebra of $V^{*}$: this can be important doing graded algebra computations).

1. For all $e_i\in\wedge^1 V$ and $w_j\in\wedge^{1} V^{*}$

   $i_{w_j}(e_i):=w_j(e_i)$.

2. For all $e_{i_1}\wedge\dots\wedge e_{i_k}\in\wedge^k V$ and $w_j\in\wedge^{1} V^{*}$

$i_{w_j}(e_{i_1}\wedge\dots\wedge e_{i_k}):=\sum_{l=1}^k(-1)^{l-1}e_{i_1}\wedge\dots\wedge w_j(e_l)\wedge\dots\wedge e_{i_k}$.

The $(-1)^{l-1}$ appears because we are moving $w_j$ "through" $e_{i_1}\wedge\dots\wedge e_{i_{l-1}}$ to contract $w_j$ with $e_{i_l}$.

1. $e_{i_1}\wedge\dots\wedge e_{i_k}\in\wedge^k V$ and $w_{j_1}\wedge\dots\wedge w_{j_n}\in\wedge^n V^{*}$

$i_{w_{j_1}\wedge\dots\wedge w_{j_n}}(e_{i_1}\wedge\dots\wedge e_{i_k}):= i_{w_{j_1}}(i_{w_{j_2}}(\dots (i_{w_{j_n}}(e_{i_1}\wedge\dots\wedge e_{i_k})))$.

I hope this helps

Answer 2

How about recursively? $\alpha$ can be decomposed as an exterior product of $k$ covectors $\alpha_1, \alpha_2, \ldots, \alpha_k$. Then $i_\alpha = i_{\alpha_1} i_{\alpha_2} \ldots i_{\alpha_k}$. Perhaps in that order or perhaps reversed. There might be some kind of sign convention in play (revering the order of the products), but this should be the gist of it.

Answer 3

Ok I expand my comment.

Take $e^*_1 \in \bigwedge^k(V^*)$, $e_1 \wedge e_2 \in \bigwedge^n(V)$

Than $$i_{e*_1}(e_1\wedge e_2) = e^*_1(e_1)\wedge e_2 - e_1 \wedge e^*_1(e_2) = e_2$$

And now take: $\frac{1}{2}e^*_1 \in \bigwedge^k(V^*)$, $2 e_1 \wedge e_2 \in \bigwedge^n(V)$

Observe that you get the same result:

$$i_{\frac{1}{2}e*_1}(2 e_1\wedge e_2) = 2\frac{1}{2}e^*_1(e_1)\wedge e_2 - 2\frac{1}{2}e_1 \wedge e^*_1(e_2) = e_2$$

Note: Not sure if I have a signs right. If I write $i_a^*(b\wedge c) = a^*(b)\wedge c - b\wedge a^*(c)$. It might be $i_a^*(b\wedge c) = - a^*(b)\wedge c b\wedge a^*(c)$ cant remember what the sign convention is.

So the $(\alpha,\sigma) \rightarrow i_\alpha \sigma$ cannot be injection as you state it.

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We cleared things up. So we want to show that for any nonzero $\sigma \in \bigwedge^n(V)$ the is isomorphism $\bigwedge^k(V^*)\cong\bigwedge^{n-k}(V)$. Given by $\alpha \in \bigwedge^k(V^*) \rightarrow i_\alpha \sigma \in \bigwedge^{n-k}(V)$.

I will not go through every step but I will show what this statement says geometricaly.

The $i_\alpha \sigma$ is linear in $\alpha$ so we show that there is one-to-one correspondence between basis.

and suppose $\sigma = e_1\wedge \dots \wedge e_n$

Example: $i_{e^*_1\wedge e^*_2} = e_3\wedge \dots \wedge e_n$(here I might be missing minus sign, can't remember) So you can see that on right hand side you get those basis vectors missing on the left hand side. I left you to fully formulate this statement in general and prove it.

I want to talk what does it mean geometrically: $e_1\wedge e_2$ can be visualized ad subspace spanned by $e_1,e_2$ and $e_3\wedge \dots \wedge e_n$ represents its orthogonal complement.

In general there is some subtlety to it. Not every element $\alpha \in \bigwedge^k(V)$ can by visualized as $k$ dimensional subspace but those if form $\alpha = v_1 \wedge \dots \wedge v_k$ represents $\mathbf{k}$ dimensional subspace spanned by $v_1,\dots,v_k$. So than $i_\alpha \sigma$ represents its orthogonal complement. ThusS it works for every element of basis.

For example $e_1\wedge e_2 + e_3\wedge e_4$ does not represents any $2$ dimensional space. Actually I had a question about this here.