# Young diagram for standard representation of $S_d$

I'm working through Fulton-Harris and I'm kind of "stuck" at the following question. I'm looking for representations of $S_d$, the symmetric group on $d$ letters via Young Tableaux. The question is: "Show that for general $d$, the standard representation $V$ of $S_d$ corresponds to the partition $d = (d-1)+1$."

When I look at the hint I see that they give a basis: $$v_j = \sum_{g(d)=j} e_g - \sum_{h(1)=j} e_h$$ for $j=2,...,d$. Is there any way I can "see" that this should be a basis corresponding to the partition? Because, right now, they just seem to appear from magic. The representation should be the image of $c_{(d-1,1)} = a_{(d-1,1)} \cdot b_{(d-1,1)}$ in the group algebra of $S_d$ (here $a_\lambda$ corresponds to the permutations of $S_d$ that preserves the rows of the young tableaux and $b_\lambda$ the columns).

I do see how one might try for smaller cases by simply multiplying each element by $c_\lambda$ ( The young symmetrizer) , but even for $d=4$, this becomes quite impractical. Are there any other ways to see that this "should be" a basis? In general, what can we say about a young diagram and the corresponding basis for its irreducible representation?

• Try to explain what all this is about precisely. What is "this basis" for instance. What group is being represented? Formulating this way it is difficult to help you out. Mar 10, 2012 at 21:16
• Thanks for the input Mark, I tried to clarify it a bit. Is it easier to understand now? Mar 10, 2012 at 22:09

The formula does not come from magic. Here's how I think about it. Suppose you give the Young diagram corresponding to the partition $$d = (d-1) + 1$$ the standard tableaux of $$1,2,3,\ldots, d-1$$ on the first row and $$d$$ for the second row. Now instead of applying the young symmetrizer straight up to $$\Bbb{C}[S_d]$$ you can first apply $$a_\lambda$$ and see what happens. Now we claim that $$\Bbb{C}[S_d]a_\lambda$$ is $$d$$ dimensional. To see this, first notice that that the row group $$P_\lambda \cong S_{d-1}$$.

Now for any $$e_g,e_h$$ with $$g,h\in S_{d-1}$$, we have $$e_g a_\lambda = e_ha_\lambda$$. This is because $$a_\lambda$$ is the sum of all elements in $$P_\lambda\cong S_{d-1}$$ and multiplying again by an $$e_g$$ for $$g \in S_{d-1}$$ just permutes the order of summation in $$a_\lambda$$. More generally, we see that for any $$e_g, e_h \in \Bbb{C}[S_d]$$ such that $$g^{-1}h \in S_{d-1}$$, we have

$$e_ga_\lambda = e_ha_\lambda.$$

This comes down to the fact that two left cosets $$gS_{d-1}$$ and $$hS_{d-1}$$ are equal iff $$g^{-1}h \in S_{d-1}$$. Hence $$\Bbb{C}[S_d]a_\lambda$$ has basis vectors $$v_i$$ that are

$$v_i = e_\sigma a_\lambda$$

where $$\sigma$$ is a 2-cycle of form $$(i\hspace{1mm} d)$$ for $$1 \leq i \leq d$$ with the convention that $$(d\hspace{1mm} d)$$ is the identity.

The final step in the problem is to apply $$b_\lambda$$ to each of these basis vectors and show that their total sum is zero. Indeed, this can be seen as follows. We write $$\big(\sum_{i=1}^d v_i\big)b_\lambda=\big(\sum_{i=1}^d v_i\big)\big(1-(1\ d)\big)$$ as

$$\begin{array}{ccccccc}\bigg(e_{(1)} a_\lambda &+& e_{(1d)}a_\lambda &+& e_{(2d)}a_\lambda &+& \ldots &+& e_{(d-1 \hspace{1mm} d)} a_\lambda \bigg) \\ &&&& \text{minus} &&\\ \bigg(e_{(1)} a_\lambda e_{(1d)} &+& e_{(1d)}a_\lambda e_{(1d)} &+& e_{(2d)}a_\lambda e_{(1d)} &+& \ldots &+& e_{(d-1 \hspace{1mm} d)} a_\lambda e_{(1d)}\bigg).\end{array}$$

Notice we can decompose $$e_{(1)}a_\lambda$$ as

$$e_{(1d)} \left(\sum_{ g\in P_\lambda, g(1) = 1} e_g \right) e_{(1d)} + e_{(1d)} \left(\sum_{ g\in P_\lambda, g(2) = 1 } e_g \right) e_{(2d)} + \ldots + e_{(1d)} \left(\sum_{ g\in P_\lambda, g(d-1) = 1 } e_g \right) e_{(d-1\hspace{1mm} d)}.$$

A similar decomposition exists for elements in the first row in that big fat expression we wrote for $$\sum_{i=1}^d v_ib_\lambda$$. You should be able to see now that the sum is in fact zero, so that $$\Bbb{C}[S_d]c_\lambda$$ is spanned by

$$v_2b_\lambda,\hspace{1mm} v_3b_\lambda, \hspace{1mm} \ldots, v_db_\lambda.$$

However each of these vectors is precisely the $$v_j$$ that they have in the answer at the back of Fulton and Harris so we are done.

Although the other answer is correct, I found it a bit confusing. In this answer I'll provide a step-by-step explanation of the proof written by @user382668. I'll denote $$e_g$$ by simply $$g$$ and $$(d - 1, 1)$$ by $$\lambda$$.

Here it goes:

• Let $$w_i = (i \; d)a_\lambda$$. First note that $$\{w_i\}_{i = 1}^d$$ is a basis for $$\mathbb{C}[S_d]a_\lambda$$ (check the other answer for further details).
• Let $$v_i = w_i b_\lambda$$. Since $$\{w_i\}_{i = 1}^d$$ is a basis for $$\mathbb{C}[S_d]a_\lambda$$, $$\{v_i\}_{i = 1}^d$$ generates $$\mathbb{C}[S_d]a_\lambda b_\lambda = V_\lambda$$.
• Note that $$\sum_{i = 1}^d v_i = 0$$, i.e. $$v_1 = - \sum_{i = 2}^d v_i$$ (check the other answer for further details). This means $$\{v_i\}_{i = 2}^d$$ generates $$V_\lambda$$.
• It follows from the Hook length formula that $$\dim V_\lambda = d - 1 = |\{v_i\}_{i = 2}^d|$$, so $$\{v_i\}_{i = 2}^d$$ is a basis for $$V_\lambda$$.
• Finally, note that $$v_i = \sum_{g(d) = i} g - \sum_{h(1) = i} h$$. In other words, $$\{v_i\}_{i = 2}^d$$ is precisely the basis in the back of Fulton-Harris. We are done.

Here's a proof of the last item: $$\begin{split} v_i & = (i\;d) a_\lambda b_\lambda \\ & = (i\;d) \left( \sum_{p \in P_\lambda} p \right) \left( \sum_{q \in Q_\lambda} \operatorname{sgn}(q) q \right) \\ & = (i\;d) \left( \sum_{g(d) = d} g \right) (1 - (1\;d)) \\ & = \sum_{g(d) = d} (i\;d)g - \sum_{h(d) = d}(i\;d)h(1\;d) \\ & = \sum_{g(d) = i} g - \sum_{h(1) = i} h \end{split}$$

On the first answer, I couldn't see $$\sum_{i=1}^{d} v_i = 0$$ using the decompositions. I think how we see $$\sum_{i=1}^{d} v_i = 0$$ from them should be explained.

So I will prove it with the second answer's proof.

Following the second answer, $$\forall i, v_i = \sum_{g(d)=i} g - \sum_{h(1)=i} h$$ holds. Hence, when we think the sum of all $$v_i$$, \begin{align*} \sum_{i=1}^{d} v_{i} &= \sum_{i=1}^{d} \Big( \sum_{g(d)=i} g - \sum_{h(1)=i} h \Big) \\ &= \sum_{i=1}^{d} \Big( \sum_{g(d)=i} g \Big) - \sum_{i=1}^{d} \Big( \sum_{h(1)=i} h \Big) \\ &= \sum_{g \in S_d} g - \sum_{h \in S_d} h = 0. \end{align*} Therefore, we obtain $$\sum_{i=1}^{d} v_i = 0$$.