I am reading a paper ,,The Gelfand map and symmetric product" by V.M. Buchstaber and E.G. Rees


On the page 6 in the proof of Theorem 2.8 there is considered a subspace $S^n A\subset A^{\otimes n}$ of symmetric tensors. It is written that a typical element of $a\in S^nA$ is of the form $$\textbf{a}=\sum\limits_{\sigma\in S_n}a_{\sigma(1)}\otimes\ldots\otimes a_{\sigma(n)}. $$

I can't understand what $a_{\sigma(i)}$ is and why this equality holds. Thank you very much in advance for any help in understanding this.


I never read that paper, but I am almost sure about the meaning of $$S^nA\subset A^{\otimes n}.$$

First of all: $A^{\otimes n}=A\otimes ... \otimes A$ $n$ times; so an arbitrarily element of $A^{\otimes n}$ can be write in the following way:

$$ \textbf{a}=\sum_{finite} a_1 \otimes ... \otimes a_n,$$ where $a_i \in A$ for every $i=1,...,n$.

Now $S^nA$ in the subset of symmetric elements in $A^{\otimes n}$; what does symmetric element means?

Consider the action of the group $S_n$ on $A^{\otimes n}$, which act in the following way: let $\sigma \in S_n$ and $\textbf{a}\in A^{\otimes n}$; then

$$\sigma . \textbf{a}=\sigma.(a_1\otimes...\otimes a_n)=a_{\sigma(1)}\otimes...\otimes a_{\sigma(n)}.$$

In other words every elements in $S_n$ act on an element $\textbf{a}$ permuting the elements $a_i$.

DEFINITION: An element $\textbf{a}$ is called symmetric if: for every $\sigma \in S_n$ the following equality holds: $$\sigma.\textbf{a}=\textbf{a};$$ and we say that $\textbf{a}\in S^nA$.

Finally is not difficult to see that $\textbf{a}=\sum_{\sigma \in S_n} a_{\sigma(1)}\otimes...\otimes a_{\sigma(n)}\in S^nA$.

Try to see what happens when an element of $S_n$ acts on $\sum_{\sigma \in S_n} a_{\sigma(1)}\otimes...\otimes a_{\sigma(n)}$.

  • 2
    $\begingroup$ Your description of an arbitrary element in the tensor power is not right. It is a sum of such elements. $\endgroup$ – Tobias Kildetoft Mar 2 '16 at 12:18
  • 1
    $\begingroup$ Actually, your last part is also not right. Sums of such elements need not themselves have that form. The paper in question said "typical" element which is just a lose term for "usually we just need to work with such" $\endgroup$ – Tobias Kildetoft Mar 2 '16 at 12:24
  • $\begingroup$ @TobiasKildetoft thanks for your right comments! $\endgroup$ – InsideOut Mar 2 '16 at 13:46

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