Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I am working out Fulton-Harris's Representation Theory text. For $S_3$, there is the standard representation $V$ which is two dimensional. That's all great and fine. Let $Sym^k(V)$ be the symmetric powers of $V$. I really have no good understanding of what this means, but I have been trying to power through it since I couldn't find anything good online that wasn't already super advanced. Anyways, from my ability to pick up patterns, I am guessing the that the $\chi_{Sym^k(V)}(g) = \displaystyle\sum_{i_1\leq \ldots \leq i_n} \lambda_{i_1}\cdots\lambda_{i_n}$. Since $\chi$ is a class function we break things into three cases. The class of the identity is easy, $\chi(e) = n+1$ always, since $\dim(Sym^n(V))=n+1$ (I think). In the cases of $(12)$ and $(123)$ we will want to remember that $\lambda^2=1$ and $\lambda^3$ respectively since these $S_3$ is a finite group. I believe an important tool in this computation are the complete homogeneous symmetric polynomials, but not sure yet. I have worked out the cases where $n=1, 2, 3, 4$ (not sure if I should do $n=0$):

$n=1$ This is just $V$ so the character is the same as that of $V$

$n=2$ We have $\lambda_1^2+\lambda_1\lambda_2+\lambda_2^2$ which we can write as $\frac12 \chi(g)^2+\chi(g^2)$ since $\chi(g)^2=(\lambda_1+\lambda_2)^2=\lambda_1^2+2\lambda_1\lambda_2+\lambda_2^2$.

Similarly for $n=3$, We have $\lambda_1^3+\lambda_1^2\lambda_2+\lambda_1\lambda_2^2+\lambda_2^3$ which we can write as $\frac13 \chi(g)^3+2\chi(g^3)$ since $\chi(g)^3=(\lambda_1+\lambda_2)^3=\lambda_1^3+3\lambda_1^2\lambda_2+3\lambda_1\lambda_2^2+\lambda_2^3$.

However, when $n=4$, things get hairy: $\lambda_1^4+\lambda_1^3\lambda_2+\lambda_1^2\lambda_2^2+\lambda_1\lambda_2^3+\lambda_4$ Let's start with $\chi(g)^4=(\lambda_1+\lambda_2)^4= \lambda_1^4+4\lambda_1^3\lambda_2+6\lambda_1^2\lambda_2^2+4\lambda_1\lambda_2^3+\lambda_4$. In the case where $g=(12)$ we have $3+2\lambda_1\lambda_2=3+\frac{\chi(g)^2-\chi(g^2)}2$ and in the case where $g=(123)$ we have $\lambda_1+\lambda_2+\lambda_1^2\lambda_2^+\lambda_1+\lambda_2=2\lambda_1+2\lambda_2+\lambda_1^2+\lambda_2^2$ Not sure where to go from here and I have no clue how to generalize this for all $n$.

Please help! Am I on the right track?

share|cite|improve this question
Just out of curiosity, before I attempt a response--are you having difficulty understanding what $\mathcal{S}^k(V)$ means, or what the representations look like? – Alex Youcis Apr 2 '12 at 4:48
@AlexYoucis: Both. I read that the definition of a symmetric $k^th$ power of an arbitrary vector space with basis $\{e_i\}$ is the vector space with basis $\{e_{i_1}\cdots e_{i_k}\}$ where $i_1 \leq \ldots i_k$. – Steven-Owen Apr 2 '12 at 4:59
The easiest way to think about symmetric powers is to imagine your vector space to consist of linear functions on some other vector space (indeed you can take that other space to be the dual $V^*$ of $V$, but the important thing is just to think of your vectors as functions). Then $\mathrm{Sym}^k(V)$ consists of all (homogeneous) polynomial functions obtained by multiplying together $k$ of those linear functions at a time, and adding such terms together. The "symmetric" here comes from the commutativity of multiplication. All distinct products of $k$ basis functionals gives a basis of Sym. – Marc van Leeuwen Apr 2 '12 at 8:10
up vote 4 down vote accepted

I'm putting this as a response, because I can't fit into the comment section.

Understanding the definition of the $k^{\text{th}}$ symmetric power is probably more important at this point than understanding the representation theory. Thus, I'll try to address the definition of the symmetric powers and once you have taken that in either I or another member will help you with the representation theory.

I'll try to give a you a quick overview, but suggest that you take a look at a book like Dummit and Foote. Basically the idea for the symmetric algebra over $V$ is to build a sort of "polynomial ring" where one incorporates the preexisting vector space structure of $V$.

How exactly should one go about doing this? There are generally three ways, all of which coincide since we are (by my own choice) dealing with $K$-vector spaces where $K$ is characteristic zero (if you don't know what this means, just take $K=\mathbb{C}$).

I will try to explain the one which is most common as low-tech as possible--if it doesn't help I, or another member, can try a different approach.

This explanation requires you to understand the tensor algebra of the vector space $V$.

Ok, so to build the tensor algebra on $V$ one defines $T^k(V)=V^{\otimes k}$ with, by convention, $T^0(V)=K$. When then defines the vector space $T(V)$ to be $\displaystyle \bigoplus_{k\geqslant 0}T^k(V)$. So an element of $T(V)$ will look something like $v_1\otimes v_2+3(w_1\otimes w_2\otimes w_3\otimes w_4)+\alpha$ where $\alpha\in K$ (in other words, just think about it as formal sums of tensors of different degrees). This already has a predefined vector space structure--we just formally add things, and we multiply a sum by a scalar by just multiplying each term in the sum (this will have more rigorous grounding if you think about the tensor algebra as the direct sum indicated). The question then is now how one defines a product (for this is desirable). One can define the product of two simple tensors (of any degree) in $T(V)$ by 'concatenation' namely we take an element of $T^k(V)$ and an element $T^\ell(V)$ and multiply them to get an element of $T^{k+\ell}(V)$

$$(v_1\otimes\cdots\otimes v_k)\cdot(w_1\otimes\cdots\otimes w_\ell)=v_1\otimes\cdots\otimes v_k\otimes w_1\otimes\cdots\otimes w_\ell$$

This completely formality in the multiplication, the inability to simplify, is what makes the tensor algebra so "general".

Ok, now that we have ring structure on $T(V)$ we can discuss ideals and the resulting quotient rings. But, lurking in the background is more than just a ring structure. Indeed, we have a "grading" of $T(V)$. Think about it like polynomials--for each $k$ we have a feeling for the elements of "gradation" $k$ namely the elements of $T^k(V)$ which are just sums of simple tensors of length $k$. This is important because it is where the $k$ in $S^k(V)$ will come from. Ok, so what does this grading have to do with ideals? Well, we can take the set of all elements $X=\left\{v\otimes w-w\otimes v:v,w\in V\right\}\subseteq T^2(V)$ and consider the ideal $I=\langle X\rangle$ (the ideal generated by $X$). How the grading plays into this is that (as I leave to you to check) $\displaystyle I=\sum_{k}I_k$ where $I_k=I\cap T^k(V)$. Thus, the ideal itself has a natural grading. Thus, when we take the quotient algebra (ring/vector space) $T(V)/I\overset{\text{def.}}{=}$ we should expect that it has a grading of its own. Indeed, one can check that $T(V)/I$ is naturally graded as

$$T(V)/I=\bigoplus_k T^k(V)/I_k$$

We then define $S^k(V)$ to be equal to this $T^k(V)/I_k$.

So, what exactly does $S(V)$, and more pertinently, $S^k(V)$ look like? Well, by quotienting $T(V)$ by $I$ what we really did was take a multiplication that was totally free of relations (no simplifications can be made) and turned it into a ring where the multiplication is almost free of relations--but now we require that the multiplication is commutative. So, for example where previously $v_1\otimes v_2\otimes v_3\ne v_2\otimes v_1\otimes v_3$ (in general) this now must hold (to see why this is "multiplicative" think about it as $v_1\cdot v_2\cdot v_3=v_2\cdot v_1\cdot v_3$ with our concatenation multiplication). Ok, fine, so that's what $S(V)$ looks like, it just looks like $T(V)$ where now multiplication is commutative (a polynomial ring over $V$ in a sense)--but what does $S^k(V)$ look like? Well, intuitively all we did by restricting ourselves to $S^k$ was to force commutativity in $T^k(V)$ (where pedagogically it's helpful to think of, again, $v_1\otimes\cdots\otimes v_k=v_1\cdots v_k$). Of course, how we actually force commutativity is that we now have to think of $v_1\otimes\cdots\otimes v_k$ (or sums of simple tensors of length $k$) s cosets $\overline{v_1\otimes\cdots\otimes v_k}$ where we can simplify tensors by rearranging them any which way we want--which, of course, manifests itself explicitly by relations such as $\overline{v_1\otimes\cdots\otimes v_k}=\overline{v_2\otimes v_1\otimes\cdots\otimes v_k}$.

I hope that made sense, it was hard to keep track of everything!

share|cite|improve this answer
$I$ is not the union but the sum of the $I_k$. In the definition of concatenation, a ... is missing. $v_2\cdot v_2\cdot v_3$ should be $v_2\cdot v_1\cdot v_3$. – darij grinberg Apr 2 '12 at 18:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.