Show via Young diagrams that the standard representation of $S_d$ corresponds to the partition $d=(d-1)+1$

Question

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?

Answer 1

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.

Answer 2

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: $$ \begin{split} v_i & = (i\;d) a_\lambda b_\lambda \\ & = (i\;d) \left( \sum_{p \in P_\lambda} p \right) \left( \sum_{q \in Q_\lambda} \operatorname{sgn}(q) q \right) \\ & = (i\;d) \left( \sum_{g(d) = d} g \right) (1 - (1\;d)) \\ & = \sum_{g(d) = d} (i\;d)g - \sum_{h(d) = d}(i\;d)h(1\;d) \\ & = \sum_{g(d) = i} g - \sum_{h(1) = i} h \end{split} $$

Answer 3

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 $\sum_{i=1}^{d} v_i = 0$.