# The mapping to alternating tensors

1. Jun 17, 2011

### yifli

I'm wondering why $1/k!$ is needed in Alt(T), which is defined as:
$$\frac{1}{k!}\sum_{\sigma \in S_k} \mbox{sgn}\sigma T(v_{\sigma(1)},\cdots,v_{\sigma(k)})$$

After removing $1/k!$, the new $\mbox{Alt}$, $\overline{\mbox{Alt}}$, still satisfies $\overline{\mbox{Alt}}(T)(v_1,\cdots,v_i,\cdots,v_j,\cdots,v_k)=-\overline{\mbox{Alt}}(T)(v_1,\cdots,v_j,\cdots,v_i,\cdots,v_k)$, which means $\overline{\mbox{Alt}}$ is an alternating tensor

2. Jun 17, 2011

### micromass

Staff Emeritus
If T is alternating, then Alt(T)=T. This is not true with your new Alt.
Also, Alt(Alt(T))=Alt(T) does not remain true for your new Alt.

3. Jun 18, 2011

### mathwonk

But if you do not care about those properties, i.e. if you just want a map and not a "projection", then you do not need it.

Indeed there is a sense in which this is artificial. I.e. the alternating tensor product is more naturally a quotient module than a submodule of the tensor product, and these properties do not make sense there.

I.e. the space of "tensors" on a vector space V, is really the dual of the tensor product of V with itself, and the space of alternating tensors is really the dual of the alternating tensor product. Moreover the dual of the tensor product is isomorphic to the tensor product of the dual, and the same for the alternating products.

But there is a completely natural "projection" from the tensor product of the duals to its quotient, the alternating product, and this natural projection does not correspond to the one above with the 1/k! in it.

I may be confused about this as it has been a long time, but it interested me a s a student and I worked it out this much. People writing about tensors as multilinear or alternating functions, are using the dual approach and sometimes may not know the abstract "tensor product of modules" approach. (Spivak does know it however, and apparently just chooses which property he likes better in writing about this topic.)