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Suppose two sets of covectors on a vector space $V$, $\beta^1,\ldots,\beta^k$ and $\gamma^1,\ldots,\gamma^k$, are related by $$\beta^i=\sum_{i=1}^ka^i_j\gamma^i,\quad i=1,…,k,$$ for a $k\times k$ matrix $A=[a^i_j]$. Show that $$\beta^1 \wedge\cdots\wedge\beta^k=(\det A)\gamma^1\wedge\cdots\wedge\gamma^k.$$ This is a problem on Tu's textbook "introduction to manifolds" (problem 3.7). I've been working on this, and I just don't seem to understand what to do. When I tried to write out the definitions for the wedge product, everything just seemed to get worse.

If someone would please offer some helpful hints, I would appreciate it. I'm not trying to cheat, I'd really prefer to understand all of this material.

Pires Dankan
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anon271828
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  • I suggest looking at the uniqueness properties of the determinant. – Micah Jan 29 '13 at 21:31
  • @Micah: What do you mean? Whenever I tried to write out the wedge product, I wasn't sure what to do with the coefficients of the $\gamma^i$, but it didn't seem like there was anything in the book that was similar. – anon271828 Jan 29 '13 at 21:32
  • The determinant is the unique real-valued function on square matrices that's both antisymmetric and multilinear (up to a scaling factor). So you can solve this problem by showing that: 1) the wedge product of the $\beta$s is always a scalar multiple of the wedge product of the $\gamma$s, and 2) that the function sending $[a^i_j]$ to this scalar multiple is multilinear, antisymmetric, and $1$ on the identity matrix. The advantage of this approach is that, if you set it up right, you don't really have to do much calculation.... – Micah Jan 29 '13 at 21:44

2 Answers2

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I realize this is an old question, but I was looking at it for homework and thought I would share with you how I solved it. I felt as if the purpose of the exercise was for the reader to explicitly compute the value as the previous exercises in the text were proving the wedge is alternating and linear.

\begin{align*} \beta^1\wedge\cdots\wedge\beta^k&=\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)\\ &=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}\\ &=\sum_{\sigma\in S_k}\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{\sigma(1)}\wedge\cdots\wedge \gamma^{\sigma(k)}\\ &=\sum_{\sigma\in S_k}\text{sgn}\sigma\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{1}\wedge\cdots\wedge \gamma^{k}\\ &=(\det A)\gamma^{1}\wedge\cdots\wedge \gamma^{k} \end{align*}

Here the first equality holds from the fact that the wedge product is linear and antisymmetric, so you just distribute and zero out terms and you'll be left with all permutations of $1,\ldots,k$ of gamma. Then you pull out the constants and rearrange your gammas. Rearranging brings out the sign of the permutation, then you are left with what you desire.

hurloon
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This is a well-known result. It becomes a bit clearer in index-free notation for linear operators. Let's start with two covectors.

$$\beta^i \wedge \beta^j = \underline a(\gamma^i) \wedge \underline a(\gamma^j)$$

This can be taken as a definition of $\underline a(\gamma^i \wedge \gamma^j)$, and it can be extended to larger wedge products.

When the manifold has dimension $k$, then $\gamma^1 \wedge \ldots \wedge \gamma^k$ is the pseudoscalar of the manifold. The highest-ranked wedge product of vectors forms a one-dimensional vector space---all other wedge products of $k$ (co)vectors form only scalar multiples of the pseudoscalar.

Hence, it becomes clear that, for some scalar $\alpha$,

$$\beta^1 \wedge \beta^2 \wedge \ldots \wedge \beta^k = \underline a(\gamma^1 \wedge \gamma^2 \wedge \ldots \wedge \gamma^k) = \alpha \gamma^1 \wedge \gamma^2 \wedge \ldots \wedge \gamma^k$$

This construction fulfills the needs Micah describes. The conclusion is that $\alpha = \det \underline a$. It's certainly a lot simpler than trying to prove things with indices.

Muphrid
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  • Sorry I just got out of a class. What do you mean by $\underline{a}$? Another question is what do you mean by "highest-ranked wedge product" because I've not heard this terminology before? – anon271828 Jan 30 '13 at 00:09
  • $\underline a$ is the linear operator that transforms $\gamma^i$ to $\beta^i$. The underline is just to denote it as such an operator. Objects formed by wedge products of several vectors typically have a rank or grade that describes their dimension. A vector is grade 1. A wedge product of two vectors is grade 2, and so on. If the manifold has dimension $k$, then the pseudoscalar has grade $k$---and no other objects of higher grade can be formed because of the antisymmetry properties of the wedge product. – Muphrid Jan 30 '13 at 00:14