The kernel of an alternator $\pi:L^k(V) \to L^k(V)$

abstract-algebramultilinear-algebratensors

Define a funtion $$\begin{align*}
\pi:L^k(V) \to L^k(V)
\end{align*}$$ I call $\pi$ is an alternator.

$L^k(V)$ denotes the space of $k$ tensors on vector space $V$ and
$$
\pi(f) = \sum_{\sigma} \operatorname{sgn}(\sigma) \, \sigma(f)
$$
where $\sigma$ denotes all possible permutations from
$S_k = \{1,\ldots,k\}$ to $S_k$ itself and
$$
\sigma(f)(v_1,\dots,v_k)=f(v_{\sigma(1)},\dots,v_{\sigma(k)})
$$
where $v_1,\dots,v_k$ are vectors from $V$ .

I want to find the kernel of map $\pi$ . As so far, I can know that if $f$ is a symmetric tensor, then it should be in the kernel of $\pi$. Let $\tau$ be changing $i \to j, \ j \to i$ and remain other positions, we have
\begin{align}
\pi(\tau(f))
&= \sum_{\sigma} \operatorname{sgn} (\tau^{-1} \circ \tau \circ \sigma) (\tau \circ \sigma) f \\
&= \text{sgn } (\tau^{-1}) \sum_{\sigma}(\tau \circ \sigma) (\tau \circ \sigma) f \\
&= \text{sgn } (\tau) \pi(f) \\
&= – \pi (f)
\end{align}

On the other hand, since $f$ is symmetric, so
$$
\tau(f) = f
$$

Thus we have
$$
\pi(f) = -\pi(f) \implies \pi(f) = 0
$$

However, I think the kernel of $\pi$ should be all symmetric $k$ tensors, I got stuck when try to prove
$$
\pi(f) = 0 \implies f \text{ is symmetric}
$$

Any help on this? Thanks

Btw: I only learned multilinear algebra and tensors through Munkres' Analysis on Manifolds and it doesn't talk about the decomposition of tensors and very advanced knowledge of multilinear algebra (at least so far). Thus, if it's possible, please explain in detail. Thanks

Best Answer

First note that if $k=1$, $\pi$ is the identity map and $\ker \pi = 0$. So, suppose from here that $k>1$.

Definition. A tensor $f \in \mathcal{L}^k(V)$ is redundant if $f = \phi_1 \otimes \cdots \otimes \phi_k$ for $\phi_1,\dots,\phi_k \in \mathcal{L}^1(V)$ and $\phi_i = \phi_{i+1}$ for some $1 \leq i < k$.

Observe that if $e_i$ is the elementary permutation that exchanges $i$ and $i+1$, then $e_if = f$ and, since $\pi(e_i f) = \text{sgn}(e_i) \pi(f) = -\pi(f)$, so $\pi(f)=0$.

Definition. Let $\mathcal{I}^k(V)$ be the spanned subspace of $\mathcal{L}^k(V)$ by all redundant $k$-tensors.

Hence, $\ker \pi$ contains $\mathcal{I}^k(V)$. We will prove that they are indeed equal.

Lemma. Given $f \in \mathcal{L}^k(V)$, for any $\sigma \in S_k$ there exists $g_\sigma \in \mathcal{I}^k(V)$ such that $$ \sigma f = \text{sgn}(\sigma)f+g_\sigma. $$ Proof: Since $\sigma$ can be written as a product of elementary permutations, we will prove the statement by induction on the number of factors.

  • Suppose $\sigma=e_i$ for some $i$, and assume without loss of generality that $f$ equals $\phi_1 \otimes \cdots \otimes \phi_k$ for $\phi_1,\dots,\phi_k \in \mathcal{L}^1(V)$. Then $$ \sigma f-\text{sgn}(\sigma)f = f_1 \otimes (\phi_i \otimes \phi_{i+1} + \phi_{i+1} \otimes \phi_i) \otimes f_2, $$ where $f_1 = \phi_1 \otimes \cdots \otimes \phi_{i-2}$ and $f_2 = \phi_{i+2} \otimes \cdots \otimes \phi_k$. Notice that the tensor $\phi_i \otimes \phi_{i+1} + \phi_{i+1} \otimes \phi_i$ belongs to $\mathcal{I}^2(V)$ since it can be written as $$ \tfrac12 [(\phi_i + \phi_{i+1}) \otimes (\phi_i + \phi_{i+1}) - \phi_i \otimes \phi_i - \phi_{i+1} \otimes \phi_{i+1}]. $$ It follows that $\sigma f-\text{sgn}(\sigma)f \in \mathcal{I}^k(V)$.
  • For the inductive step, write $\sigma$ as $e \circ \tau$, where $e$ is an elementary permutation and $\tau$ is a product containing at least one such permutation. Then, by the previous case and the hypothesis, $$ g_1 := e(\tau f) - \text{sgn}(e)(\tau f) \in \mathcal{I}^k(V), \\ g_2 := \tau f - \text{sgn}(\tau)f \in \mathcal{I}^k(V). $$ Hence \begin{align} \sigma f = e(\tau f) &= \text{sgn}(e)(\tau f) + g_1 \\ &= \text{sgn}(e)(\text{sgn}(\tau)f + g_2) + g_1 \\ &= \text{sgn}(\sigma)f + (\text{sgn}(e)g_2+g_1). \end{align}

Proposition. For any $f \in \mathcal{L}^k(V)$ there exists $g \in \mathcal{I}^k(V)$ such that $$ \pi(f) = k!f+g. $$

Proof. Indeed: \begin{align} \pi(f) &= \sum_{\sigma \in S_k} \text{sgn}(\sigma)(\sigma f) \\ &= \sum_{\sigma \in S_k} (f+\text{sgn}(\sigma)g_\sigma) = k!f + \sum_{\sigma \in S_k} \text{sgn}(\sigma)g_\sigma. \end{align}

Corollary. The kernel of $\pi$ is $\mathcal{I}^k(V)$.

Proof: If $f \in \ker \pi$ and $g \in \mathcal{I}^k(V)$ is such that $(0=) \pi(f) = k!f+g$, then $$ f = -\frac1{k!}g \in \mathcal{I}^k(V). $$

Reference:

Victor Guillemin and Peter J. Haine, Differential Forms. Section 1.5 (page 17).

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