> [!definition] > > A $k$**-[[Tensor|tensor]]** $\omega \in \mathcal{J}^k(V)$ is **alternating** if $\omega \circ \sigma = \sgn(\sigma) \cdot \omega$ for all $\sigma \in S_k$. > [!definition] > > The set $\Lambda^k(V) \subset \mathcal J^k(V)$ is the collection of all alternating $k$-tensors. > [!definition] > > Let $T \in \mathcal J^k(V)$, define > $ > \text{Alt}(T) = \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot T \circ \sigma > $ > then > 1. $\text{Alt}(T) \in \Lambda^k(V)$. > 2. If $T \in \Lambda^k(V)$, then $\text{Alt}(T) = T$. > > *Proof*. Let $\tau \in S_k$, then the map $S_k \to S_k$ with $\sigma \to \sigma\tau$ is a bijection. > $ > \begin{align*} > \text{Alt}(T) \circ \tau &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot T \circ \sigma \circ \tau \\ > &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot T \circ \sigma\tau \\ > &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma\tau^{-1}) \cdot T \circ \sigma \\ > &= \sgn(\tau^{-1}) \cdot \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot T \circ \sigma \\ > &= \sgn(\tau) \cdot \text{Alt}(T) > \end{align*} > $ > Suppose that $T \in \Lambda^k(V)$, then > $ > \begin{align*} > \text{Alt}(T) &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot T \circ \sigma \\ > &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k} \sgn(\sigma) \cdot \sgn(\sigma) \cdot T \\ > &= \frac{1}{\abs{S_k}} \sum_{\sigma \in S_k}T = T > \end{align*} > $ > [!theorem] > > The set of all > $ > \bigwedge_{j = 1}^{k}\phi_{i_j} \quad \bracs{i_j}_1^k \text{ strictly increasing} > $ > is a basis for $\Lambda^k(V)$, which therefore has dimension ${n \choose k}$. > > *Proof*. If $T \in \Lambda^k(V) \subset \mathcal{J}^k(V)$, then > $ > w = \sum_{i \in [n]^k}a_i \bigotimes_{j = 1}^{k}\phi_{i_j} > $ > and > $ > T = \text{Alt}(T) = \sum_{i \in [n]^k}a_i \text{Alt}\paren{\bigotimes_{j = 1}^{k}\phi_{i_j}} > $ > Since each $\text{Alt}\paren{\bigotimes_{j = 1}^{k}\phi_{i_j}}$ is a constant multiple of $\bigwedge_{j = 1}^k\phi_{i_j}$, these elements span $\Lambda^k(V)$. > [!theorem] > > Let $\list{v}{n}$ be a basis of $V$, $T \in \Lambda^n(V)$, and $A \in F^{n \times n}$ be a matrix. If $w_i = \sum_{j = 1}^{n}Av_j$ are $n$ vectors in $V$, then > $ > T(w_1, \cdots, w_n) = \det A \cdot T(v_1, \cdots, v_n) > $ > *Proof*. Let $S \in \mathcal{J}^n(F^n)$ by > $ > S(A) = \omega(Av_1, \cdots, Av_n) > $ > then $S$ is alternating, and is equal to $\lambda \cdot \det$ for some $\lambda$, and $\lambda = S(e_1, \cdots, e_n) = T(v_1, \cdots, v_n)$.