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Let $V$ be a finite-dimensional vector space over the field $F$ with basis $\mathcal B = \{v_1,\dots,v_n\}$. Let $1\leq k\leq n$ and pick some $1\leq i_1 < i_2 < \dots < i_k \leq n$. I am defining the $F$-linear map $\Phi\colon \mathcal T^k(V)\to F$ by its action on the basis vectors of $\mathcal T^k(V)$: $\Phi(v_{j_1}\otimes v_{j_2}\otimes\dots\otimes v_{j_k}) = \epsilon(\sigma)$ if $\sigma$ is the unique permutation of $(j_1,j_2,\dots,j_k)$ into $(i_1,i_2,\dots,i_k)$, and $\Phi$ is zero on every basis vector whose $k$-tuple of indices cannot be permuted to $(i_1,i_2,\dots,i_k)$ (where $\epsilon(\sigma)$ is the sign of $\sigma$). By the universal property of the tensor product, there is a unique map $\phi\colon V^k\to F$ that is $k$-multilinear over $F$ such that $\phi(w_1,w_2,\dots,w_k) = \Phi(w_1\otimes w_2\otimes\dots\otimes w_k)$.

I would like to show $\phi$ is alternating. Let $(w_1,\dots,w,w,\dots,w_k)\in V^k$. I know that $$ \phi(w_1,\dots,w,w,\dots,w_k) = \Phi(w_1\otimes\dots\otimes w\otimes w\otimes\dots\otimes w_k). $$ But how can I show that this is zero? I think I should write $w_1\otimes\dots\otimes w\otimes w\otimes\dots\otimes w_k$ as a linear combination of basis vectors of $\mathcal T^k$; I know that those vectors in the linear combination whose indices are not a permutation of $(i_1,i_2,\dots,i_k)$ will get mapped to zero under $\Phi$, but I need to show that the images of the remaining summands (i.e., those whose indices are permutations of $(i_1,i_2,\dots,i_k)$) cancel each other out. I think it has something to do with the sign of $\epsilon(\sigma)$ but I'm at a bit of a loss.

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Instead of showing that $\Phi(w_1\otimes \cdots\otimes w\otimes w\otimes \cdots\otimes w_k)$ vanishes, let's show that $$\Phi(w_1\otimes \cdots\otimes w_i\otimes w_{i+1}\otimes \cdots\otimes w_k)=-\Phi(w_1\otimes \cdots\otimes w_{i+1}\otimes w_{i}\otimes \cdots\otimes w_k)$$

(Away from characteristic $2$, these conditions are equivalent.)

Indeed, if $\sigma$ is the permutation which puts the indices of the first tensor in increasing order, then $\sigma\circ (i,\; i +1)$ is the permutation corresponding to the second tensor. Since the sign is multiplicative and $\epsilon((i,\; i+1))=-1$, the result follows.

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