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

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 $i_{\alpha}\sigma$?

• Are you sure that the ismorphism is between $\bigwedge^k(V^*)\otimes\bigwedge^n(V)$ and $\bigwedge^{n-k}(V)$. As I can remember from class the statement was that for every nonzero element of $\bigwedge^n(V)$ there is isomorphism between $\bigwedge^k(V^*)$ and $\bigwedge^{n-k}(V)$. – tom Jun 13 '13 at 8:12
• Yes and the wiki page is in accordance with my claim. They say that interior product $\bigwedge^k(V^*)\otimes\bigwedge^n(V)\cong\bigwedge^{n-k}(V)$ only induce the isomorphism. – tom Jun 13 '13 at 8:28

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

• Do you have a reference for this? – Michael Albanese Jun 17 '13 at 9:27
• To be honest...no :-) Surely some good information is contained in the classical "Multilinear Algebra" by Greub. Personally I wrote these formulae when dealing with Koszul complexes. – Avitus Jun 17 '13 at 9:58
• Don't you want to reverse the order of contractions on the right hand side of the last line? As written, you'd have $i_{w_1\wedge w_2}(e_1\wedge e_2) = i_{w_1}(i_{w_2}(e_1\wedge e_2)) = i_{w_1}(-e_1) = -i_{w_1}(e_1) = -1$, but surely you want $i_{w_1\wedge w_2}(e_1\wedge e_2) = 1$ right? – Michael Albanese Jun 17 '13 at 11:28
• With my convention $\wedge V$ becomes a left $\wedge V^{*}$ module, as $i_{w \wedge r}( e_{i_1}\wedge\dots\wedge e_{i_l}):= i_{w}(i_{r}(e_{i_1}\wedge\dots\wedge e_{i_l})),$ Reversing the order probably induces non trivial signs, as we exchange 1-forms. – Avitus Jun 17 '13 at 11:51
• Either way, you get an isomorphism. I suppose it just depends on the context. – Michael Albanese Jun 17 '13 at 12:02

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.

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.

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.

• In your example, you are considering the map $(\alpha, \sigma) \mapsto i_{\alpha}\sigma$, but the map in question is $\alpha\otimes\sigma \mapsto i_{\alpha}\sigma$. Note that $$\frac{1}{2}e^*_1\otimes(2e_1\wedge e_2) = e^*_1\otimes(e_1\wedge e_2)$$ so you argument against injectivity fails. – Michael Albanese Jun 16 '13 at 4:43
• Ahh sorry I didn't realized this. – tom Jun 16 '13 at 8:11