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$S_n$ acts on $V^{\otimes n}$ ($V$: a vector space) from the right as $$x_1\otimes ...\otimes x_n.\sigma=x_{\sigma(1)}\otimes ...\otimes x_{\sigma(n)},$$ but I don't see why as I get $$x_1\otimes ...\otimes x_n.(\sigma\circ\tau)=x_{\sigma(\tau(1))}\otimes ...\otimes x_{\sigma(\tau(n))}$$ which is not equal to $$(x_1\otimes ...\otimes x_n.\sigma).\tau=x_{\sigma(1)}\otimes ...\otimes x_{\sigma(n)}.\tau=x_{\tau(\sigma(1))}\otimes...\otimes x_{\tau(\sigma(n))}.$$

So what am I doing wrong here?

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    You are switching $\sigma$ and $\tau$ when you write $\sigma\circ \tau$. Remember, $\circ$ means after. So $\sigma\circ \tau$ is $\tau$ first and then $\sigma$, exactly the opposite of what you wanted. In other words, the multiplication in $S_n$ is defined as follows: $\sigma\cdot \tau= \tau\circ \sigma$, using this everything is fine :) – Mathematician 42 Jan 23 '18 at 11:13
  • Honestly I don't understand, the product is what it is, the $\circ$. So acting with $\sigma\tau$ means first changing the order with $\tau$, then with $\sigma$ which is not the same as changing first with $\sigma$, then with $\tau$. –  Jan 23 '18 at 11:43
  • Think about this: How do you define the product in $S_n$? Is it $\sigma\cdot \tau:= \sigma\circ \tau$ or $\sigma\cdot \tau:=\tau\circ \sigma$? Which of these products makes the above an action? – Mathematician 42 Jan 23 '18 at 12:32
  • Alternatively, you could avoid the confusion by writing the permutations consistently on the right: $j\sigma$, not $\sigma(j)$. – ancient mathematician Jan 24 '18 at 11:30

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To avoid confusion on how $\tau$ acts on $(x_1\otimes ...\otimes x_n.\sigma)$, you could rename $v_i = x_{\sigma(i)}$ and notice that $(v_1\otimes ...\otimes v_n.\tau)=v_{\tau(1)}\otimes ...\otimes v_{\tau(n)} = x_{\sigma(\tau(1))}\otimes ...\otimes x_{\sigma(\tau(n))}$.

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