Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say I want to put action of $S_4$ on $V^{\otimes 4}$. If I want to act on the left, why can't I say

$$\sigma(v_1 \otimes v_2 \otimes v_3 \otimes v_4) = v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(4)}?$$

Everything seems to work out right e.g. take $\sigma = (123),\tau = (134)$. Then $\sigma\tau = 234$.

$$(\sigma\tau)(v_1 \otimes \ldots \otimes v_4) = v_1 \otimes v_3 \otimes v_4 \otimes v_2$$

and $\sigma\big( \tau(v_1 \otimes \ldots \otimes v_4) \big)$ is also the same thing. But books I see say we have to throw in an inverse for the left action? I am getting confused because it seems throwing the inverse does not make it an action! What's happening?

share|improve this question
    
@leo They define $\sigma(v_1 \otimes \ldots v_4) = v_{\sigma^{-1}(1)} \otimes \ldots v_{\sigma^{-1}(4)}$. But how is this a left action? Associativity fails! –  user23086 Oct 14 '12 at 3:01
    
Associativity does not fail. –  Qiaochu Yuan Oct 14 '12 at 3:14
    
@QiaochuYuan Take $\tau = (13)$ and $\sigma = 12$ in $S_4$. Then if I go by the left action using inverses, $(\tau\sigma)(v_1 \otimes \ldots \otimes v_4) = v_3 \otimes v_1 \otimes v_2 \otimes v_4$. However $\tau( \sigma(v_1 \otimes \ldots \otimes v_4)) = \tau(v_2 \otimes v_1 \otimes v_3 \otimes v_4) =v_2 \otimes v_3 \otimes v_1 \otimes v_4$. –  user23086 Oct 14 '12 at 3:14
    
Your computations are incorrect. See my answer. –  Qiaochu Yuan Oct 14 '12 at 3:14

1 Answer 1

up vote 4 down vote accepted

You aren't computing $\sigma(\tau(v_1 \otimes ... \otimes v_4))$ correctly. We have

$$\tau(v_1 \otimes ... \otimes v_4) = v_3 \otimes v_2 \otimes v_4 \otimes v_1.$$

Now rename $w_1 = v_3, w_2 = v_2, w_3 = v_4, w_4 = v_1$. Then

$$\sigma(w_1 \otimes ... \otimes w_4) = w_2 \otimes w_3 \otimes w_1 \otimes w_4$$

which is equal to

$$v_2 \otimes v_4 \otimes v_3 \otimes v_1$$

and so is not equal to $(\sigma \tau)(v_1 \otimes ... \otimes v_4)$. It is, however, equal to $(\tau \sigma)(v_1 \otimes ... \otimes v_4)$.

When you apply $\sigma$ after applying $\tau$, $\sigma$ doesn't remember what indices the $v_i$ had originally. It only sees their current positions, and then it permutes those.

share|improve this answer
    
Thanks very much! –  user23086 Oct 14 '12 at 3:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.