I'm going to use @ to denote tensor product.
1. If pi is a permutation of the numbers 1 through n, and VxVx...xV is the set of n-tuples (v_1,v_2,...,v_n), define pi.(v_1,v_2,...,v_n) to be (v_{pi(1)}, v_{pi(2)}, ..., v_{pi(n)}), rearranging the n-tuple.
Since VxVx...xV is the basis for V*V*....*V, we can linearly extend this definition to V*V*...*V. Show that each pi preserves the subspace I of V*V*...*V. Therefore, pi gives a well-defined map from V @ V @...V.
A. There are two kinds of generators of I; those that move a scalar in and out of an entry, and those that move addition in and out of an entry. If you apply a pi to one of these generators, you get another one, except that the scalar/+ may be moving in or out of a different entry.
If we have some element of the tensor product, and lift it up to V*V*...*V, we make a choice, and different choices differ by elements of I. When we apply pi to it, pi(the different choices) differ by elements of pi(I). But we now know that's I. So when we map down to V@V@V...@V, the choices wash out.
2. If V is n-dimensional, define an isomorphism of V @ V with nxn matrices, in such a way that "apply the permutation 1<->2" (from question 1) turns into "transpose".
A. Pick a basis of V, which allows us to isomorph V with row and also with column vectors; call these maps v |-> row(v), v |-> col(v).
Map V*V -> nxn matrices by taking (v,w) to col(v) times row(w). That's obviously bilinear in v,w, so kills I, so descends to the tensor product.
Then switching (v,w) to (w,v) switches col(v) row(w) to col(w)row(v), which is the transpose.
3. Define Sym^2 V = { t in V@V: (1<->2).t = t }. What's its dimension?
A. By the previous question, we can think in terms of matrices, and ask about the matrices that are equal to their transposes. These are the symmetric matrices. To write one down, we specify the upper triangle, which is n(n+1)/2 numbers.
4. Define the map Alternate : V@V@...@V to itself by
Define Alt^n V to be the image of this map, inside V@V@...@V (n factors).
Given a basis v_1...v_d of V, define an "ordered pure tensor" as one of the form v_(i_1) @ v_(i_2) @ .. @ v_(i_d), where i_1 < i_2 < ... < i_d.
Let R be the subspace of V@V@...@V spanned by the ordered pure tensors. Show that Alternate : R -> Alt^n V is an isomorphism.
A. First let's show that it's onto.
We have a nice basis of the tensor product: _all_ pure tensors of basis elements. If we can write each of these as a combination of ordered pure tensors plus something in ker(Alternate), then we'll know the restriction to R is still onto.
First check that if a pure tensor has a repeated vector, like a @ b @ a @ c, then the Alternate of it is zero. This is because we can match up the terms in which the a's are switched to the ones in which they're not, and the signs are off by one switch so -1. Each pair cancels.
Another fine thing to find in the kernel is the sum of two pure tensors that differ by switching two things, like a@b@c@D@e@F@g + a@b@c@F@e@D@g. The argument is exactly the same as above.
We can use this to reorder basis elements, for example
All together, this lets us write any pure tensor of basis elements as either an element of the kernel, or an element of the kernel plus something in R. So R maps onto the image.
Now we have to show it's 1:1. Let t be a linear combination of ordered pure tensors, and let m be the lexicographically first one in the sum. (We can only say this if t is nonzero.) When we hit t with a permutation, the terms we get are always later than m lexicographically. Adding them up with signs, we get Alternate(t), which still has m as its lex-first term. In particular, Alternate(t) is not zero.
5. What's the dimension of Alt^n V?
We just gave a basis of (d choose n) elements, for an isomorphic space.
6. If T : V->W, find a natural map from Alt^n V -> Alt^n W.
A. First, let's define such a map from V@V@V@...@V -> W@W@...@W: