Symmetrizing and Anti-Symmetrizing Tensors

Given any Tensor, we can obtain a symmetric tensor through symmetrising operator. by

$T_{uv} \rightarrow T_{(uv)}=\frac{1}{n!}(T_{uv}+T_{vu})$ where $n$ is the order of the tensor, and you have to take up all the permutations of the indices possible.

And for anti-symmterising tensors, you have to take up alternating sums of the per mutated tensor components, and you get an anti-symmteric tensor.

Now so this is a map from a tensor space to itself.

Is this map only a change of basis map ? Is this a one-one map ?

It is true given any $$p$$-covariant (or $$p$$ contravariant, that is not important) tensor $$T$$ you can define two new tensors:

First, the Symmetric tensor $$T_S$$ given in coordinates by $$(T_S)_{i_1,\dots,i_p} = \frac{1}{p!} \sum_{\sigma \in \mathfrak{S}_p} T_{i_{\sigma(1)} , \dots i_{\sigma(p)}} \, ,$$ where $$\mathfrak{S}_p$$ is the symmetric group of order $$p$$ and $$i_j$$ runs from 1 to $$n= \mbox{dim}(E)$$, ($$E$$ the vector sapce considered), $$j=1,2,\dots, p$$.

And also the skew-symmetric tensor $$T_A$$ ($$A$$ from antisymmetric), in coordinates reading $$(T_A)_{i_1,\dots,i_p} = \frac{1}{p!} \sum_{\sigma \in \mathfrak{S}_p} \mbox{sgn}(\sigma) \, T_{i_{\sigma(1)} , \dots i_{\sigma(p)}} .$$

So you are right, but your formulae are wrong.

For $$p=2$$ you recover the usual matrix decomposition. However, @Andrew D. Hwang is right: for $$p>2$$ the usual formula fails (and if I am not worng you need some representation theory to find the correct decomposition).

And both spaces cannot be one-to-one, because their dimensions are different. They already are in the usual case $$p=2$$, when skew-symmetric tensors have dimension $$n(n-1)/2$$ and the symmetric ones have dimension $$n(n+1)/2$$. For arbitrary $$p$$, of the space of all $$p$$-covariant (contravariant, remember the construction is the same) skew-symmetric tensors for is the binomial coefficient $$n$$ over $$p$$ if $$p; and zero for $$p>n$$.

Finally, notice that the skew-symmetrisation of every symmetric tensor will be the zero-tensor.

It is true however that, if your vector space has an inner product, the spaces of $$p$$-skew-symmetric tensors and $$(n-p)$$-skew-symmetric tensors are isomorphic. The isomorphism is called Hodge dual operator (also Hodge star operator or Hodge dual star operator). (An isomorphism exists even if you do not have an inner product, since two finite-dimensional vector spaces with the same dimension are isomorphic).

Skew-symmetric covariant tensors are very important in differential geometry. You can find information about that seeing exterior algebra, exterior calculus, $$p$$-form...

I hope my answer helps you.

Not totally sure about your notation ("$uv$" represents an $n$-tuple and the sum represents summing over all permutations?), but viewed as operators on $\bigotimes^{n}(V^{*})$, the space of all $n$-multilinear functions on $V$, neither symmetrization nor anti-symmetrization is one-one or onto.

For $n > 2$, the images of these maps do not even span $\bigotimes^{n}(V^{*})$, i.e., not every $n$-tensor is the sum of a symmetric and anti-symmetric tensor.