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I found this question on the site and the first answer which states that

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

and I'm wondering about the equalities in this. Firstly how is $$\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right) = \sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}$$

and secondly how is

$$\sum_{\sigma\in S_k}\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{\sigma(1)}\wedge\cdots\wedge \gamma^{\sigma(k)} = \sum_{\sigma\in S_k}\text{sgn}\sigma\left(\prod_{i=1}^ka^i_{\sigma(i)}\right)\gamma^{1}\wedge\cdots\wedge \gamma^{k}?$$

There seems to be some usage of some property related to the wedge product here which I'm not familiar with. I know the definition, but this seems to be not using that and instead something else?

Rivaldo
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  • Learn about the map $Alt$ which sends a tensor to an alternating tensor, and is scaled so that when you use it to define the wedge product , the wedge product is associative. Everything will fall in place. I recommend Loren Tu’s introduction to manifolds. – Charlie Frohman Jan 18 '23 at 20:01

2 Answers2

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  1. The first step is actually two steps in one. The wedge product is multi-linear, so we expand $$\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\dotsc\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)=\sum_{j_1,\dotsc,j_k=1}^ka^1_{j_1}\gamma^{j_1}\wedge\dotsc\wedge a_{j_n}^k\gamma^{j_k}.$$ Next, the wedge product is alternating, so only the summands for which $j_1,\dotsc,j_k$ are pairwise distinct are non-zero. To say that $j_1,\dotsc,j_k$ are pairwise distinct, however, is just to say that they are a permutation of $\{1,\dotsc,k\}$, so that's how we get $$\sum_{j_1,\dotsc,j_k=1}^ka^1_{j_1}\gamma^{j_1}\wedge\dotsc\wedge a_{j_n}^k\gamma^{j_k}=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}.$$

  2. The wedge product is alternating, hence anti-symmetric, meaning $\gamma^{\sigma(1)}\wedge\dotsc\wedge\gamma^{\sigma(k)}=\mathrm{sgn}(\sigma)\gamma^1\wedge\dotsc\wedge\gamma^k$.

Thorgott
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  • I think my issue is with understanding that $j_i, \dots j_k$ pairwise distinct implies that they are a permutation of ${1, \dots, k}$. How is this true? @Thorgott – Rivaldo Jan 18 '23 at 20:18
  • @Rivaldo How do you choose $k$ numbers from $1$ to $k$ (inclusive) with no repetitions? – Ted Shifrin Jan 18 '23 at 22:54
  • @TedShifrin I'm trying to tie this up, but I have some difficulties. The group $S_k$ has $k!$ elements so the lhs of $$\sum_{j_1,\dotsc,j_k=1}^ka^1_{j_1}\gamma^{j_1}\wedge\dotsc\wedge a_{j_n}^k\gamma^{j_k}=\sum_{\sigma\in S_k}a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a^k_{\sigma(k)}\gamma^{\sigma(k)}$$ has $k!$ summands and it would seem that the rhs has much more (how many exactly?), but most of them cancel out. I don't currently see why summing over the permutations keeps only the summands that do not cancel. – Rivaldo Jan 18 '23 at 23:04
  • It's nothing as fancy as canceling. The LHS has $k^k$ summands, the RHS has $k!$ summands. Perhaps think of it like this: a tuple $(j_1,\dotsc,j_k)$ with $j_1,\dotsc,j_k\in{1,\dotsc,k}$ is the same thing as a map ${1,\dotsc,k}\rightarrow{1,\dotsc,k}$. The elements $j_1,\dotsc,j_k$ are pairwise distinct if and only if this corresponding map is a bijection, i.e. a permutation of ${1,\dotsc,k}$. All the summands on the LHS corresponding to tuples of elements that are not pairwise distinct vanish, so only the summands on the RHS remain. – Thorgott Jan 18 '23 at 23:13
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    @Thorgott I think I got it now from your comment. In a sense I'm picking tuples $(j_1, \dots, j_k)$ and the problematic ones which end up vanishing are those where I for example have $j_n = j_m$ for distinct $n$ and $m$ (perhaps more than $2$). Therefore I'm only able to pick tuples where all the $j_i$'s are distinct which would correspond to this permutation you described. – Rivaldo Jan 18 '23 at 23:23
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$$\begin{align*} \beta^1\wedge\cdots\wedge\beta^k&=\left(\sum_{j=1}^ka_j^1\gamma^j\right)\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right) \\ &=\sum_{j=1}^k \left(a_j^1\gamma^j\wedge\cdots\wedge\left(\sum_{j=1}^ka_j^k\gamma^j\right)\right) \\ &= \cdots \\ &= \sum_{1 \leq i_1, \ldots, i_k \leq k} \left(a_{i_1}^1\gamma^{i_1}\wedge\cdots\wedge a_{i_k}^k\gamma^{i_k}\right) \\ &=\sum_{\sigma \in S_k} \left(a_{\sigma(1)}^1\gamma^{\sigma(1)}\wedge\cdots\wedge a_{\sigma(k)}^k\gamma^{\sigma(k)}\right), \end{align*} $$ where the last equality comes up because we can't have repeats (by antisymmetry).

Now let $\sigma \in S_k$. If $\sigma = (13)$, then notice that

$$\begin{align*} \gamma^3 \wedge \gamma^2 \wedge \gamma^1 \wedge \cdots \wedge \gamma^k &= -\gamma^3 \wedge \gamma^1 \wedge \gamma^2 \wedge \cdots \wedge \gamma^k \\ &= \gamma^1 \wedge \gamma^3 \wedge \gamma^2 \wedge \cdots \wedge \gamma^k \\ &= -\gamma^1 \wedge \gamma^2 \wedge \gamma^3 \wedge \cdots \wedge \gamma^k .\end{align*}$$

Check that this works for every transposition. Using the definition of $\text{sgn}$, the second equality follows.

User203940
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