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Let $V$ be a vector space over $\mathbb{R}$. A $p$-tensor on $V$ is a multilinear map $$ \omega: \underbrace{V\times\cdots\times V}_\text{$p$ times}\longrightarrow\mathbb{R} $$ The set of all $p$-tensors is denoted $\mathcal{T}^p(V)$.

Let $\omega\in\mathcal{T}^p(V)$, and define $A:\mathcal{T}^p(V)\longrightarrow \Lambda^p(V)$ by $$ A(\omega)(v_1,\ldots,v_p) = \frac{1}{p!}\sum_{\sigma\in S_p}{(-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})} $$ Finally, let $\omega\in\mathcal{T}^p(V)$ and $\eta\in\mathcal{T}^q(V)$, and define the tensor product $\omega\otimes\eta\in\mathcal{T}^{p+q}(V)$ $$ \omega\otimes\eta(v_1,\ldots,v_p,v_{p+1},\ldots,v_{p+q}) = \omega(v_1,\ldots,v_p)\eta(v_{p+1},\ldots,v_{p+q}) $$

I'm trying to prove the following identity: $$ A(A(\omega\otimes\eta)\otimes\nu) = A(\omega\otimes A(\eta\otimes\nu)) = A(\omega\otimes\eta\otimes\nu) $$ Here was my initial approach:

Let $l=p+q+r$ and $m=p+q$. \begin{align*} A(A(\omega\otimes\eta)&\otimes\nu)(v_1,\ldots,v_l) = \frac{1}{l!}\sum_{\rho\in S_{l}}{(-1)^\rho A(\omega\otimes\eta)(v_{\rho(1)},\ldots,v_{\rho(m)})\nu(v_{\rho(m+1)},\ldots, v_{\rho(l)})}\\ &= \frac{1}{l!m!}\sum_{\rho\in S_{l}}{\sum_{\sigma\in S_m}(-1)^{\sigma\rho}\omega(v_{\sigma\rho(1)},\ldots,v_{\sigma\rho(p)})\eta(v_{\sigma\rho(p+1)},\ldots,v_{\sigma\rho(m)})\nu(v_{\rho(m+1)},\ldots, v_{\rho(l)})}\, . \end{align*}

Let $H\subset S_l$ be a set of all permutations such that for any $g,h\in H$, $g(i) = h(i)$ is not true for all $i=m,\ldots,l$. Then, any element of $S_l$ can be written as $\rho h$ for some $\rho\in S_m$. This is equivalent to representing a permutation of $S_l$ as first selecting $r$ ordered elements from $\{1,\ldots,l\}$, placing them at the end of a list and then permuting the remaining $m$ elements and placing them at the start. The total number of ways to do this is equal to $(l!(l-r)!)m!=l!=|S_l|$. Thus the above sum can be rewritten as

\begin{align*} \frac{1}{l!m!}\sum_{h\in H}(-1)^h\nu(v_{h(m+1)},\ldots,v_{h(l)})\Bigg({\sum_{\rho,\sigma\in S_m}{(-1)^{\sigma\rho h}\omega(v_{\sigma\rho h(1)},\ldots,v_{\sigma\rho h(p)})\omega(v_{\sigma\rho h(p+1)},\ldots,v_{\sigma\rho h(m)})}\Bigg)} \end{align*}

I should be able to reduce the second sum to just $$ m!\sum_{\rho\in S_m}{(-1)^{\rho}\omega(v_{\sigma\rho h(1)},\ldots,v_{\sigma\rho h(p)})\omega(v_{\sigma\rho h(p+1)},\ldots,v_{\sigma\rho h(m)})} $$ Which would give me the desired result but I can't seem to figure out how. Any help would be appreciated.

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1 Answer 1

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Firstly, you have to be a little more careful with the indices. Let $w_i = v_{\rho(i)}$ for $1\le i\le m$. Then $$ \begin{align*} \newcommand{\tens}{\otimes} A(\omega\tens\eta)(w_1,\ldots,w_m) =& \frac{1}{m!} \sum_{\sigma\in S_m} (-1)^\sigma \omega(w_{\sigma(1)},\ldots,w_{\sigma(p)}) \eta(w_{\sigma(p+1)},\ldots,w_{\sigma(m)}) \\ =& \frac{1}{m!} \sum_{\sigma\in S_m} (-1)^\sigma \omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)}) \eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)}). \end{align*} $$ Note that this results in a different order of composition of $\rho$ and $\sigma$ in the index. This is crucial.

Now we can do the following to simplify the double sum. The key steps, are noticing that for $i>m$, $\sigma(i)=i$, so $\rho\sigma(i)=\rho(i)$, giving the first equality below, and that we can then reindex by summing over $\tau = \rho\sigma$. $$ \begin{align*} \frac{1}{l!m!} \sum_{\rho\in S_{l}}& \sum_{\sigma\in S_m}(-1)^{\sigma\rho} \omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)}) \eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)}) \nu(v_{\rho(m+1)},\ldots, v_{\rho(l)}) \\ =& \frac{1}{l!m!} \sum_{\rho\in S_l} \sum_{\sigma\in S_m} (-1)^{\sigma\rho} \omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)}) \eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)}) \nu(v_{\rho\sigma(m+1)},\ldots, v_{\rho\sigma(l)}) \\ =&\newcommand{\inv}{^{-1}} \frac{1}{l!m!} \sum_{\sigma\in S_m} \sum_{\substack{\tau\in S_l\\\rho:=\tau\sigma^{-1}}} (-1)^{\sigma\tau\sigma\inv} \omega(v_{\tau\sigma\inv\sigma(1)},\ldots,v_{\tau\sigma\inv\sigma(p)}) \eta(v_{\tau\sigma\inv\sigma(p+1)},\ldots,v_{\tau\sigma\inv\sigma(m)}) \nu(v_{\tau\sigma\inv\sigma(m+1)},\ldots, v_{\tau\sigma\inv\sigma(l)}) \\ =& \frac{1}{l!m!} \sum_{\sigma\in S_m} \sum_{\tau\in S_l} (-1)^{\tau} \omega(v_{\tau(1)},\ldots,v_{\tau(p)}) \eta(v_{\tau(p+1)},\ldots,v_{\tau(m)}) \nu(v_{\tau(m+1)},\ldots, v_{\tau(l)}) \\ =& \frac{m!}{l!m!} \sum_{\tau\in S_l} (-1)^{\tau} \omega(v_{\tau(1)},\ldots,v_{\tau(p)}) \eta(v_{\tau(p+1)},\ldots,v_{\tau(m)}) \nu(v_{\tau(m+1)},\ldots, v_{\tau(l)}) \\ =& A(\omega\tens\eta\tens\nu)(v_1,\ldots,v_l) \end{align*} $$

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