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 John Armstrong | Algebra, Linear Algebra, Representation Theory

14 Comments »

[…] Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if has dimension , then the space of antisymmetric tensors with tensorands […]

Pingback by The Determinant « The Unapologetic Mathematician | December 31, 2008 | Reply
I cannot help but wonder if it makes any sense to change the factorials by Euler gamma functions and talk about spaces non-integer dimensions. Is this a field (of study) in pure mathematics?

Comment by Melvin | January 21, 2009 | Reply
Melvin, it all depends on what you mean by “dimension”. Remember that we’ve defined the dimension of a vector space as the cardinality of a basis, and cardinalities of (finite) sets are always natural numbers.

Comment by John Armstrong | January 21, 2009 | Reply
@Melvin: I guess maybe you’ve heard of fractals and fractal dimensions?

Perhaps more relevant to gamma functions per se are various analytic expressions which come up under the rubric of “dimensional regularization” in quantum field theory, where so-called “D-dimensional integrals” come into play for complex values of D; gamma factors seem to arise frequently there for example. This reminds me too that there is a notion of fractional differentiation that people sometimes talk about, where the Cauchy formula for the $n^{th}$ derivative of an analytic function,

$\displaystyle f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{|z - z_0| = r} \frac{f(z)}{(z - z_0)^{n+1}} dz$

is extended to non-integer values of $n$ , with an obvious substitution of the Gamma function for $n!$ .

Tangential but fun fact: the $(n-1)$-dimensional volume of the unit ball in $\mathbb{R}^n$ is $\pi^{n/2}/(n/2)!$ , interpreted as appropriate for both odd and even $n$ . This I feel deserves a non-factorial exclamation mark 🙂

Comment by Todd Trimble | January 23, 2009 | Reply
John: I agree that cardinality is a natural number. The point is if one can make any sense of vector spaces with positive non-integer dimension. You can ignore me, I am sure this daydream is getting me a lot of points in the crackpot index. 😉

Todd: Dimensional regularization was what triggered my daydream and then I found this nice post on vector spaces. More precisely, in QFT (at least for QED) one does not usually bother with changing the dimensionality of spinors and one keeps it appropriate to four spacetime dimensions. The momentum integrals are lifted into arbitrary dimensions and we physicist do what we do. In class I heard a remark about how Dirac spinors do not generalize in a straightforward way to arbitrary number of dimensions; even for the D = 2,3,4,5,6,7,8,9 modulo 8 cases each one is very different.

My daydream was something like what if one could use some framework where all the objects have a nice generalization to arbitrary dimensions and some physical requirement (like what happens in string theory) fixes the dimensionality to be a positive but real number. It would be like the ultimate joke from Nature. (Again, crackpot index running high.)

I am sure all I am saying reduces to taking the expression “analytic continuation” to seriously. After all dimensional regularization is an intermediate step in the calculation and is just one particular pick of a regularization scheme. Good physics does not depend on the choice of this scheme.

Fractional calculus and fractals sound interesting and I certainly know nothing about that. At some point I will check it out.

Comment by M. E. Irizarry-Gelpí | January 23, 2009 | Reply
Just one more random thought then. I don’t know whether you (Melvin) follow the writings of John Baez, for example his This Week’s Finds, but speaking of cardinalities, there is an interesting idea he’s written about here and there where one considers the cardinality not just of sets but more generally of groupoids, which can be fractional. I’ll curb the impulse to say much about it in a compressed comment, but check it out. I think you’d find a lot worth ruminating on, coming as you do from a physics perspective.

Comment by Todd Trimble | January 23, 2009 | Reply
Thanks for the pointer!

Comment by M. E. Irizarry-Gelpí | January 23, 2009 | Reply
I’ll point out this as a direction to ponder: what would it mean to have a groupoid as a basis, rather than just a set?

To be even heavier-handed: how is a set s groupoid? If you’ve been reading since I was doing category theory you should be able to tell.

Comment by John Armstrong | January 23, 2009 | Reply
It’s a rather mysterious (at least, to me) fact that sometimes goes by the name of “combinatorial reciprocity” that the dimension of the symmetric tensor space can also be written $(-1)^n {-d \choose n}$ . James Propp and others have done some interesting work exploring a way to make “negative cardinality” rigorous, and it’s interesting to think about how to combine these ideas with groupoid ideas to get both positive and negative real cardinalities.

Comment by Qiaochu Yuan | June 7, 2009 | Reply
My gut feeling, Qiaochu (and you may have considered this yourself) is that this type of thing can be understood formally via superalgebra (i.e., tensor calculus in $\mathbb{Z}_2$ -vector spaces). The rough idea here is that exterior powers (with dimension d choose n) are the “supersymmetric counterparts” of symmetric powers. In fact, my preferred stopgap technique for dealing with negative cardinalities is via superalgebra.

If you’d like, we can discuss this a little: I have an email account topologicalmusings@gmail.com.

Comment by Todd Trimble | June 8, 2009 | Reply
Argh, I meant $\mathbb{Z}_2$ –graded vector spaces.

Comment by Todd Trimble | June 8, 2009 | Reply
[…] is represented by an antisymmetric tensor with the sides as its tensorands. But we’ve calculated the dimension of the space of such tensors: . That is, once we represent these parallelepipeds by antisymmetric […]

Pingback by Parallelepipeds and Volumes II « The Unapologetic Mathematician | November 3, 2009 | Reply
[…] just keep going and let our methods apply to such more general tensors. Anyhow, we also know how to count the dimension of the space of such […]

Pingback by The Hodge Star « The Unapologetic Mathematician | November 9, 2009 | Reply
[…] can I eliminate non-symmetric the cases? I found this but I could not get the intution. Also, this blog post answers my question but I don’t see why we put | between different indices. Any concrete […]

Pingback by The Dimension of the Symmetric $k$-tensors - MathHub | April 3, 2016 | Reply

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