3
$\begingroup$

Problem

Given the tensor algebra: $$TV:=\sum_{k=0}^\infty{\bigotimes}^k V$$

Regard the symmetrization and antisymmetrization: $$S_\pm\left(\bigotimes_{i=1}^kv_i\right):=\frac{1}{k!}\sum_{\sigma\in\mathcal{S}_k}(\pm1)^{\mathrm{sgn}\sigma}\bigotimes_{i=1}^kv_{\sigma(i)}$$

Then they satisfies: $$S\pm\left(S_\pm\left(\bigotimes_{i=1}^kv_i\right)\otimes v_{k+1}\right)=S_\pm\left(\bigotimes_{i=1}^kv_i\otimes v_{k+1}\right)$$ How to prove these relations?

Attempt

After changing summations they become: $$\frac{1}{k+1!}\frac{1}{k!}\sum_{\sigma\in\mathcal{S}_{k}}\sum_{\tau\in\mathcal{S}_{k+1}}(\pm1)^{\mathrm{sgn}\sigma+\mathrm{sgn}\tau}\bigotimes_{i=1}^kv_{\tau(\sigma(i))}\otimes v_{\tau(k+1)}$$ Substituting inner permutation gives: $$\tau'_\sigma(1\leq i\leq k):=\tau(\sigma(i))\quad\tau'_\sigma(k+1):=\tau:\quad\mathrm{sgn}\tau'_\sigma=\mathrm{sgn}\sigma+\mathrm{sgn}\tau$$ Then the outer sum repeats so they reduce to the desired: $$\frac{1}{k+1!}\sum_{\tau'_\sigma\in\mathcal{S}_{k+1}}(\pm1)^{\mathrm{sgn}\tau'_\sigma}\bigotimes_{i=1}^{k+1}v_{\tau'_\sigma(i)}$$ But was to deal with the signum right?

$\endgroup$
2
$\begingroup$

There is no problem with this proof, but there is one point that requires a bit of clarification. Note that if we change the index of summation to $\tau_{\sigma}'$ we get $$\frac{1}{(k+1)!}\frac{1}{k!}\sum_{\sigma\in S_k}{\sum_{\tau_{\sigma}'\in S_{k+1}}{(\pm 1)^{\mathrm{sgn}(\tau_{\sigma}')}\bigotimes_{i=1}^{k+1}{v_{\tau'_{\sigma}(i)}}}}$$ Although ostensibly this depends on $\sigma$, all the application of $\sigma$ does is permute the order of the terms. The sign of the permutation is correct, so regardless of what $\sigma$ is the sum is the same. Thus we are summing the same thing $k!$ times, hence the result is $$\frac{1}{(k+1)!}\sum_{\tau_{\sigma}'\in S_{k+1}}{(\pm 1)^{\mathrm{sgn}(\tau_{\sigma}')}\bigotimes_{i=1}^{k+1}{v_{\tau'_{\sigma}(i)}}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.