# Tensor Algebra: Symmetrization & Antisymmetrization

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?

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 $$\frac{1}{(k+1)!}\sum_{\tau_{\sigma}'\in S_{k+1}}{(\pm 1)^{\mathrm{sgn}(\tau_{\sigma}')}\bigotimes_{i=1}^{k+1}{v_{\tau'_{\sigma}(i)}}}$$