Exercise 12.13 (Equivalent Definitions of Symmetric Tensors)

Question

Let $V$ be a finite-dimensional vector space. A covariant $k$-tensor $\alpha$ on $V$ is said to be symmetric if its value is unchanged by interchanging any pair of arguments:
$$
\alpha\left(v_1, \ldots, v_i, \ldots, v_j, \ldots, v_kight)=\alpha\left(v_1, \ldots, v_j, \ldots, v_i, \ldots, v_kight)
$$
whenever $1 \leq i<j \leq k$.
ewline

	extbf{Exercise 12.13.} Show that the following are equivalent for a covariant $k$-tensor $\alpha$ :
\begin{enumerate}[label=(\alph*)]
\item $\alpha$ is symmetric.
\item For any vectors $v_1, \ldots, v_k \in V$, the value of $\alpha\left(v_1, \ldots, v_kight)$ is unchanged when $v_1, \ldots, v_k$ are rearranged in any order.
\item The components $\alpha_{i_1 \ldots i_k}$ of $\alpha$ with respect to any basis are unchanged by any permutation of the indices.
\end{enumerate}

Elu Thingol · Accepted Answer

Recall that interchanging any pair of arguments $v_1,\ldots,v_k$ is equivalent to applying transpositions $	au$ to get 
\begin{equation*}
    v_{	au(1)},\ldots,	extcolor{blue}{v_{	au(i)}},\ldots,	extcolor{red}{v_{	au(j)}},\ldots,v_{	au(k)}
    = v_{1},\ldots,v_{i-1},	extcolor{blue}{v_{j}},v_{i+1},\ldots,v_{j-1},	extcolor{red}{v_{i}},v_{j+1},\ldots,v_{k}
\end{equation*}

Similarly, rearranging the values $v_1,\ldots,v_k$ in any order means applying a permutation $\sigma \in S_k$ to get $v_{\sigma(1)},\ldots,v_{\sigma(k)}$.

\begin{proof}
\begin{description}
    \item[(a) $\implies$ (b)] Suppose $\alpha$ is symmetric. Recall that every permutation $\sigma$ can be written as a composition of pairwise permutations $	au_i$, i.e., transpositions, which only exchange two elements. Since transpositions of inputs do not change the value of a symmetric tensor by assumption, the assertion follows.

\item[(b) $\implies$ (c)] Suppose that for any vectors $v_1, \ldots, v_k \in V$, the value of $\alpha$ is unchanged when $v_1, \ldots, v_k$ are rearranged in any order, i.e,
    \begin{equation*}
        \alpha\left(v_1, \ldots, v_kight) = \alpha\left(v_{\sigma(1)}, \ldots, v_{\sigma(k)}ight)
    \end{equation*}
    for some permutation $\sigma \in S_k$. In particular, this applies to basis vectors $E_1,\ldots,E_k$, i.e., for all combinations $1 \leq i_1,\ldots,i_k \leq k$:
    $$\alpha_{i_1,\ldots,i_k} = \alpha(E_{i_1},\ldots,E_{i_k}) = \alpha(E_{\sigma(i_1)},\ldots,E_{\sigma(i_k})) = \alpha_{\sigma(i_1),\ldots,\sigma(i_k)}.$$

\item[(c) $\implies$ (a)] Suppose that we want to exchange the $i$th and the $j$th arguments. Using the unique coefficients $\alpha_{i_1,\ldots,i_k}$ of our covariant $k$-tensor $\alpha$, we rewrite the value as follows. Although the steps may seem trivial, we carefully write down each step.
    \begin{align*}
        &\alpha\left(v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k}ight)\
        &= \sum_{i_{1}=1}^{k} \ldots 	extcolor{blue}{\sum_{i_{i}=1}^{k}} \ldots 	extcolor{red}{\sum_{i_{j}=1}^{k}}\ldots\sum_{i_{k}=1}^{k}
        \alpha_{i_1,\ldots,	extcolor{blue}{i_i},\ldots,	extcolor{red}{i_j},\ldots,i_k} \cdot 
        \varepsilon^{i_1} \otimes \ldots \otimes 	extcolor{blue}{\varepsilon^{i_i}}\otimes\ldots\otimes	extcolor{red}{\varepsilon^{i_j}}\otimes \ldots \varepsilon^{i_k}
        (v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k})
    \end{align*}
    By assumption, $\alpha_{i_1,\ldots,	extcolor{blue}{i_i},\ldots,	extcolor{red}{i_j},\ldots,i_k} = \alpha_{i_1,\ldots,	extcolor{red}{i_j},\ldots,	extcolor{blue}{i_i},\ldots,i_k}$
    \begin{align*}
        &\alpha\left(v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k}ight)\
        &= \sum_{i_{1}=1}^{k} \ldots 	extcolor{blue}{\sum_{i_{i}=1}^{k}} \ldots 	extcolor{red}{\sum_{i_{j}=1}^{k}}\ldots\sum_{i_{k}=1}^{k}
        \alpha_{i_1,\ldots,	extcolor{red}{i_j},\ldots,	extcolor{blue}{i_i},\ldots,i_k} \cdot 
        \varepsilon^{i_1} \otimes \ldots \otimes 	extcolor{blue}{\varepsilon^{i_i}}\otimes\ldots\otimes	extcolor{red}{\varepsilon^{i_j}}\otimes \ldots \varepsilon^{i_k}
        (v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k})
    \end{align*}
    Rearrange the sums, putting $j$th variable first
    \begin{align*}
        &\alpha\left(v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k}ight)\
        &= \sum_{i_{1}=1}^{k} \ldots 	extcolor{red}{\sum_{i_{j}=1}^{k}} \ldots 	extcolor{blue}{\sum_{i_{i}=1}^{k}} \ldots\sum_{i_{k}=1}^{k}
        \alpha_{i_1,\ldots,	extcolor{red}{i_j},\ldots,	extcolor{blue}{i_i},\ldots,i_k} \cdot 
        \varepsilon^{i_1} \otimes \ldots \otimes 	extcolor{blue}{\varepsilon^{i_i}}\otimes\ldots\otimes	extcolor{red}{\varepsilon^{i_j}}\otimes \ldots \varepsilon^{i_k}
        (v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k})
    \end{align*}
    Finally, notice that $\varepsilon^{1}\otimes\varepsilon^{2}(v_1,v_2) = \varepsilon^{2}\otimes\varepsilon^{1}(v_2,v_1)$. By renaming the iteration variables $i_i$ to $i_j$ and vice versa, we obtain:
    \begin{align*}
        &\alpha\left(v_{1},\ldots,	extcolor{blue}{v_{i}},\ldots,	extcolor{red}{v_{j}},\ldots,v_{k}ight)\
        &= \sum_{i_{1}=1}^{k} \ldots 	extcolor{red}{\sum_{i_{j}=1}^{k}} \ldots 	extcolor{blue}{\sum_{i_{i}=1}^{k}} \ldots\sum_{i_{k}=1}^{k}
        \alpha_{i_1,\ldots,	extcolor{red}{i_j},\ldots,	extcolor{blue}{i_i},\ldots,i_k} \cdot 
        \varepsilon^{i_1} \otimes \ldots \otimes 	extcolor{red}{\varepsilon^{i_j}}\otimes\ldots\otimes	extcolor{blue}{\varepsilon^{i_i}}\otimes \ldots \varepsilon^{i_k}
        (v_{1},\ldots,	extcolor{red}{v_{j}},\ldots,	extcolor{blue}{v_{i}},\ldots,v_{k})\
        &= \sum_{i_{1}=1}^{k} \ldots 	extcolor{red}{\sum_{i_{i}=1}^{k}} \ldots 	extcolor{blue}{\sum_{i_{j}=1}^{k}} \ldots\sum_{i_{k}=1}^{k}
        \alpha_{i_1,\ldots,	extcolor{red}{i_i},\ldots,	extcolor{blue}{i_j},\ldots,i_k} \cdot 
        \varepsilon^{i_1} \otimes \ldots \otimes 	extcolor{red}{\varepsilon^{i_i}}\otimes\ldots\otimes	extcolor{blue}{\varepsilon^{i_j}}\otimes \ldots \varepsilon^{i_k}
        (v_{1},\ldots,	extcolor{red}{v_{j}},\ldots,	extcolor{blue}{v_{i}},\ldots,v_{k})\
        &=\alpha\left(v_{1},\ldots,	extcolor{red}{v_{j}},\ldots,	extcolor{blue}{v_{i}},\ldots,v_{k}ight)
    \end{align*}
\end{description}
\end{proof}
	extbf{Remark:} I realise that $i_i$ looks silly but by the time I noticed this, it was already too late to rename the iteration variables.

Exercise 12.13 (Equivalent Definitions of Symmetric Tensors)

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Exercise 12.13 (Equivalent Definitions of Symmetric Tensors)

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