The Unapologetic Mathematician

Mathematics for the interested outsider

Dimensions of Symmetric and Antisymmetric Tensor Spaces

We’ve laid out the spaces of symmetric and antisymmetric tensors. We even showed that if V has dimension d and a basis \{e_i\} we can set up bases for S^n(V) and A^n(V). Now let’s count how many vectors are in these bases and determine the dimensions of these spaces.

The easy one will be the antisymmetric case. Every basic antisymmetric tensor is given by antisymmetrizing an n-tuple e_{i_1}\otimes...\otimes e_{i_n} of basis vectors of V. We may as well start out with this collection in order by their indices: i_1\leq...\leq i_n. But we also know that we can’t have any repeated vectors or else the whole thing collapses. So the basis for A^n(V) consists of subsets of the basis for V. There are d basis vectors overall, and we must pick n of them. But we know how to count these. This is a number of combinations:

\displaystyle\dim\left(A^n(V)\right)=\binom{d}{n}=\frac{d!}{n!(d-n)!}

Now what about symmetric tensors? We can’t do quite the same thing, since now we can allow repetitions in our lists. Instead, what we’ll do is this: instead of just a list of basis vectors of V, consider writing the indices out in a line and drawing dividers between different indices. For example, consider th basic tensor of \left(\mathbb{F}^5\right)^{\otimes 4}: e_1\otimes e_3\otimes e_3\otimes e_4. First, it becomes the list of indices

\displaystyle1,3,3,4

Now we divide 1 from 2, 2 from 3, 3 from 4, and 4 form 5.

\displaystyle1,|,|,3,3,|,4,|

Since there are five choices of an index, there will always be four dividers. And we’ll always have four indices since we’re considering the fourth tensor power. That is, a basic symmetric tensor corresponds to a choice of which of these eight slots to put the four dividers in. More generally if V has dimension d then a basic tensor in S^n(V) has n indices separated by d-1 dividers. Then the dimension is again given by a number of combinations:

\displaystyle\dim\left(S^n(V)\right)=\binom{n+d-1}{d-1}=\frac{(n+d-1)!}{n!(d-1)!}

December 30, 2008 Posted by | Algebra, Linear Algebra, Representation Theory | 13 Comments

   

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