The Unapologetic Mathematician

Mathematics for the interested outsider

Corollaries of the Sign Lemma

The results we showed last time have a few immediate consequences we will have use of.

First, let t^\lambda and s^\mu are two Young tableaux of shapes \lambda and \mu, respectively, where \lambda\vdash n and \mu\vdash n. If \kappa_t\{s\}\neq0 — where \kappa_t is the group algebra element we’ve defined — then \lambda dominates \mu.

To see this, let b and c be two entries in the same row of s. They cannot be in the same column of t, since if they were then the swap (b\,c) would be in the column-stabilizer C_t. Then we could conclude that \kappa_t\{s\}=C_t^-\{s\}=0, which we assumed not to be the case. But if no two entries from the same row of s^\mu are in the same column of t^\lambda, the dominance lemma tells us that \lambda\trianglerighteq\mu.

Now if it turns out that \lambda=\mu it’s not surprising that \lambda\trianglerighteq\mu. Luckily in that situation we can say something interesting:

\displaystyle\kappa_t\{s\}=\pm e_t=\kappa_t\{t\}

Indeed, we must have \{s\}=\pi\{t\} for some \pi\in C_t, basically by the same reasoning that led to the dominance lemma in the first place. Indeed, the thing that would obstruct finding such a \pi is having two entries in some column of t needing to go on the same row of s, which we know doesn’t happen. And so we calculate

\displaystyle\begin{aligned}\kappa_t\{s\}&=\kappa_t\pi\{t\}\\&=\mathrm{sgn}(\pi)\kappa_t\{t\}\\&=\pm e_t\end{aligned}

Now if u\in M^\mu is any vector in the Specht module, and if t^\mu is a tableau of shape \mu, then \kappa_t u is some multiple of e_t. Indeed, we can write

\displaystyle u=\sum\limits_ic_i\{s_i\}

were the s_i are \mu-tableaux. For each one of these, we have \kappa_t\{s_i\}=\pm e_t. Thus we find

\displaystyle\kappa_tu=\sum\limits_i\pm c_ie_t

which is a multiple of e_t, as asserted.

December 31, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments

The Sign Lemma

As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let H\subseteq S_n be a subgroup, and also consider the S_n-invariant inner product on M^\lambda for which the distinct Young tabloids form an orthonormal basis.

First, if \pi\in H, then

\displaystyle\pi H^-=H^-\pi=\mathrm{sgn}(\pi)H^-

where H^- is the alternating sum of the elements of H. The proof basically runs the same as when we showed that \pi e_t=\mathrm{sgn}(\pi)e_t where t has shape (1^n).

Next, for any vectors u,v\in M^\lambda we have

\displaystyle\langle H^-u,v\rangle=\langle u,H^-v\rangle

Indeed, we can calculate

\displaystyle\begin{aligned}\langle H^-u,v\rangle&=\sum\limits_{\pi\in H}\langle\mathrm{sgn}(\pi)\pi u,v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi)\pi^{-1}v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi^{-1})\pi^{-1}v\rangle\\&=\sum\limits_{\tau\in H}\langle u,\mathrm{sgn}(\tau)\tau v\rangle\\&=\langle u,H^-v\rangle\end{aligned}

where we have used the facts that \mathrm{sgn}(\pi)=\mathrm{sgn}(\pi^{-1}), and that as \pi runs over a group, so does \tau=\pi^{-1}.

Next, if the swap (b\,c)\in H, then we have the factorization

\displaystyle H^-=k(1-(b\,c))

for some k\in\mathbb{C}[S_n]. To see this, consider the subgroup K=\{1,(b\,c)\}\subseteq H, and pick a transversal. That is, write H as a disjoint union:

\displaystyle H=\biguplus\limits_ik_iK

but then we can write the alternating sum

\displaystyle\begin{aligned}H^-&=\sum\limits_{\pi\in H}\mathrm{sgn}(\pi)\pi\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i+\mathrm{sgn}(k_i(b\,c))k_i(b\,c)\right)\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i-\mathrm{sgn}(k_i)k_i(b\,c)\right)\\&=\sum\limits_i\mathrm{sgn}(k_i)k_i\left(1-(b\,c)\right)\\&=\left(\sum\limits_i\mathrm{sgn}(k_i)k_i\right)\left(1-(b\,c)\right)\end{aligned}

as we stated.

Finally, if t is some tableau with b and c in the same row, and if the swap (b\,c)\in H, then

\displaystyle H^-\{t\}=0

Our hypothesis tells us that (b\,c)\{t\}=\{t\}. We can thus use the above factorization to write


December 29, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 6 Comments

Examples of Specht Modules

Let’s look at a few examples of Specht modules.

First, let \lambda=(n). The only polytabloid is

\displaystyle e_{\begin{array}{cccc}1&2&\cdots&n\end{array}}=\begin{array}{cccc}\cline{1-4}1&2&\cdots&n\\\cline{1-4}\end{array}

on which S_n acts trivially. And so S^\lambda is a one-dimensional space with the trivial group action. This is the only possibility anyway, since S^\lambda\subseteq M^\lambda, and we’ve seen that M^\lambda is itself a one-dimensional vector space with the trivial action of S_n.

Next, consider \lambda=(1^n) — with n parts each of size 1. This time we again have one polytabloid. We fix the Young tableau

\displaystyle t=\begin{array}{c}1\\2\\\vdots\\n\end{array}

Since every entry is in the same column, the column-stabilizer C_t is all of S_n. And so we calculate the polytabloid

\displaystyle e_t=\sum\limits_{\sigma\in S_n}\mathrm{sgn}(\sigma)\sigma\{t\}

We use our relations to calculate

\displaystyle\begin{aligned}\pi e_t&=\sum\limits_{\sigma\in S_n}\mathrm{sgn}(\sigma)\pi\sigma\{t\}\\&=\sum\limits_{\tau\in S_n}\mathrm{sgn}(\pi^{-1}\tau)\tau\{t\}\\&=\sum\limits_{\tau\in S_n}\mathrm{sgn}(\pi^{-1})\mathrm{sgn}(\tau)\tau\{t\}\\&=\mathrm{sgn}(\pi^{-1})\sum\limits_{\tau\in S_n}\mathrm{sgn}(\tau)\tau\{t\}\\&=\mathrm{sgn}(\pi)e_t\end{aligned}

We conclude that S^\lambda\subseteq M^\lambda is a one-dimensional space with the signum representation of S_n. Unlike our previous example, there is a huge difference between S^\lambda and M^\lambda; we’ve seen that M^\lambda is actually the left regular representation, which has dimension n!.

Finally, if \lambda(n-1,1), then we can take a tableau and write a tabloid


where the notation we’re using on the right is well-defined since each tabloid is uniquely identified by the single entry in the second row. Now, the polytabloid in this case is e_t=\mathbf{k}-\mathbf{i}, since the only column rearrangement is to swap i and k. It’s straightforward to see that these polytabloids span the subspace of M^\lambda where the coefficients add up to zero:

\displaystyle c_1\mathbf{1}+\dots+c_n\mathbf{n}\in S^\lambda\quad\Leftrightarrow\quad\sum\limits_{i=1}^nc_i=0

As a basis, we can pick \{\mathbf{2}-\mathbf{1},\mathbf{3}-\mathbf{1},\dots,\mathbf{n}-\mathbf{1}\}. We recognize this pattern from when we calculated the invariant subspaces of the defining representation of S_3. And indeed, M^\lambda is the defining representation of S_n, which contains S^\lambda as the analogous submodule to what we called V^\perp before.

December 28, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments

Specht Modules

Now we have everything in place to define the representations we’re interested in. For any partition \lambda, the Specht module S^\lambda is the submodule of the Young tabloid module M^\lambda spanned by the polytabloids e_t where t runs over the Young tableaux t of shape \lambda.

To see that the subspace spanned by the polytabloids is a submodule, we must see that it’s invariant under the action of S_n. We can use our relations to check this. Indeed, if e_t is a polytabloid, then \pi e_t=e_{\pi t} is another polytabloid, so the subspace spanned by the polytabloids is invariant under the action of S_n.

The most important fact about the Specht modules is that they’re cyclic. That is, we can generate one just by starting with a single vector and hitting it with all the elements in the group algebra \mathbb{C}[S_n]. Not all of the resulting vectors will be different, but among them we’ll get the whole Specht module. The term “cyclic” comes from group theory, where the cyclic groups are those from modular arithmetic, like \mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}. Many integers give the same residue class modulo n, but every residue class comes from some integer.

Anyway, in the case of Specht modules, we will show that the action of S_n can take one vector and give a whole basis for S^\lambda. Then any vector in the Specht module can be written as a sum of basis vectors, and thus as the action of some algebra element from \mathbb{C}[S_n] on our starting vector. But which starting vector will we choose? Well, any polytabloid will do. Indeed, if e_s and e_t are polytabloids, then there is some (not unique!) permutation \pi so that t=\pi s. But then e_t=e_{\pi s}=\pi e_s, and so e_t is in the S_n-orbit of e_s. Thus starting with e_s we can get to every vector in S^\lambda by the action of \mathbb{C}[S_n].

December 27, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 10 Comments

Permutations and Polytabloids

We’ve defined a bunch of objects related to polytabloids. Let’s see how they relate to permutations.

First of all, I say that

\displaystyle R_{\pi t}=\pi R_t\pi^{-1}

Indeed, what does it mean to say that \sigma\in R_{\pi t}? It means that \sigma preserves the rows of the tableau \pi t. And therefore it acts trivially on the tabloid \{\pi t\}. That is: \sigma\{\pi t\}=\{\pi t\}. But of course we know that \{\pi t\}=\pi\{t\}, and thus we rewrite \sigma\pi\{t\}=\pi\{t\}, or equivalently \pi^{-1}\sigma\pi\{t\}=\{t\}. This means that \pi^{-1}\sigma\pi\in R_t, and thus \sigma\in\pi R_t\pi^{-1}, as asserted.

Similarly, we can show that C_{\pi t}=\pi C_t\pi^{-1}. This is slightly more complicated, since the action of the column-stabilizer on a Young tabloid isn’t as straightforward as the action of the row-stabilizer. But for the moment we can imagine a column-oriented analogue of Young tabloids that lets the same proof go through. From here it should be clear that \kappa_{\pi t}=\pi\kappa_t\pi^{-1}.

Finally, I say that the polytabloid e_{\pi t} is the same as the polytabloid \pi e_t. Indeed, we compute

\displaystyle e_{\pi t}=\kappa_{\pi t}\{\pi t\}=\pi\kappa_t\pi^{-1}\pi\{t\}=\pi\kappa_t\{t\}=\pi e_t

December 24, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments


Given any collection H\subseteq S_n of permutations, we define two group algebra elements.

\displaystyle\begin{aligned}H^+&=\sum\limits_{\pi\in H}\pi\\H^-&=\sum\limits_{\pi\in H}\mathrm{sgn}(\pi)\pi\end{aligned}

Notice that H doesn’t have to be a subgroup, though it often will be. One particular case that we’ll be interested in is

\displaystyle\kappa_t=C_t^-=\sum\limits_{\pi\in C_t}\mathrm{sgn}(\pi)\pi

where C_t is the column-stabilizer of a Young tableau t. If t has columns C_1,\dots,C_k, then C_t=S_{C_1}\times\dots\times S_{C_k}. Letting \pi run over C_t is the same as letting \pi_i run over S_{C_i} for each i from 1 to k. That is,

\displaystyle\begin{aligned}\kappa_t&=\sum\limits_{\pi_1\in S_{C_1}}\dots\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_1\dots\pi_k)\pi_1\dots\pi_k\\&=\sum\limits_{\pi_1\in S_{C_1}}\dots\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_1)\dots\mathrm{sgn}(\pi_k)\pi_1\dots\pi_k\\&=\left(\sum\limits_{\pi_1\in S_{C_1}}\mathrm{sgn}(\pi_1)\pi_1\right)\dots\left(\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_k)\pi_k\right)\end{aligned}

so we have a nice factorization of this element.

Now if t is a tableau, we define the associated “polytabloid”

\displaystyle e_t=\kappa_t\{t\}

Now, as written this doesn’t really make sense. But it does if we move from just considering Young tabloids to considering the vector space of formal linear combinations of Young tabloids. This means we use Young tabloids like basis vectors and just “add” and “scalar multiply” them as if those operations made sense.

As an example, consider the tableau

\displaystyle t=\begin{array}{ccc}4&1&2\\3&5&\end{array}

Our factorization lets us write


And so we calculate

\displaystyle e_t=\begin{array}{ccc}\cline{1-3}4&1&2\\\cline{1-3}3&5&\\\cline{1-2}\end{array}-\begin{array}{ccc}\cline{1-3}3&1&2\\\cline{1-3}4&5&\\\cline{1-2}\end{array}-\begin{array}{ccc}\cline{1-3}4&5&2\\\cline{1-3}3&1&\\\cline{1-2}\end{array}+\begin{array}{ccc}\cline{1-3}3&5&2\\\cline{1-3}4&1&\\\cline{1-2}\end{array}

Now, the nice thing about e_t is that if we hit it with any permutation \pi\in C_t, we get \pi e_t=\mathrm{sgn}(\pi)e_t.

December 23, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 9 Comments

Row- and Column-Stabilizers

Every Young tableau t^\lambda with shape \lambda\vdash n gives us two subgroups of S_n, the “row-stabilizer” R_t and the “column-stabilizer” C_t. These are simple enough to define, but to write them succinctly takes a little added flexibility to our notation.

Given a set X, we’ll write S_X for the group of permutations of that set. For instance, the permutations that only mix up the elements of the set \{1,2,4\} make up S_{\{1,2,4\}}

Now, let’s say we have a tableau t with rows R_1,\dots,R_k. Any permutation that just mixes up elements of R_1 leaves all but the first row alone when acting on t. Since it leaves every element on the row where it started, we say that it stabilizes the rows of t. These permutations form the subgroup S_{R_1}. Of course, there’s nothing special about R_1 here; the subgroups S_{R_i} also stabilize the rows of t. And since entries from two different subgroups commute, we’re dealing with the direct product:

\displaystyle R_t=S_{R_1}\times\dots\times S_{R_k}

We say that R_t is the row-stabilizer subgroup, since it consists of all the permutations that leave every entry in t on the row where it started. Clearly, this is the stabilizer subgroup of the Young tabloid \{t\}.

The column-stabilizer is defined similarly. If t has columns C_1,\dots,C_l, then we define the column-stabilizer subgroup

\displaystyle C_t=S_{C_1}\times\dots\times S_{C_l}

Now column-stabilizers do act nontrivially on the tabloid \{t\}. The interaction between rearranging rows and columns of tableaux will give us the representations of S_n we’re looking for.

December 22, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 4 Comments

The Lexicographic Order on Partitions

The biggest problem with the dominance order is that it’s only a partial order. That is, there are some pairs of partitions \lambda and \mu so that neither \lambda\trianglerighteq\mu nor \mu\triangleright\lambda. We sometimes need a total order, and that’s where the lexicographic order comes in.

The lexicographic order gets its name because it’s just like the way we put words in order in the dictionary. We compare the first parts of each partition. If they’re different we use their order, while if they’re the same we break ties with the tails of the partitions. More explicitly, \lambda<\mu if for some i we have \lambda_j=\mu_j for all j<i, and \lambda_i<\mu_i.

As an example, here’s the lexicographic order on the partitions of 6:


Now, comparing this order to the dominance order, we notice that they’re almost the same. Specifically, every time \lambda\trianglerighteq\mu, we find \lambda>\mu as well.

If \lambda=\mu, then this is trivially true. But if \lambda\neq\mu there must be at least one i with \lambda_i\neq\mu_i. Let i be the first such index. Now we find that


and also


This inequality must go in this direction since \lambda\triangleright\mu. We conclude that \lambda_i>\mu_i, and so \lambda>\mu in the lexicographic order.

December 21, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | Leave a comment

The Dominance Lemma

We will have use of the following technical result about the dominance order:

Let t^\lambda and s^\mu be Young tableaux of shape \lambda and \mu, respectively. If for each row, all the entries on that row of s^\mu are in different columns of t^\lambda, then \lambda\trianglerighteq\mu. Essentially, the idea is that since all the entries on a row in s fit into different columns of t, the shape of t must be wide enough to handle that row. Not only that, but it’s wide enough to handle all of the rows of that width at once.

More explicitly, we can rearrange the columns of t so that all the entries in the first i rows of s fit into the first i rows of t. This is actually an application of the pigeonhole principle: if we have a column in t that contains i+1 elements from the first i rows of s, then look at which row each one came from. Since i+1>i, we must have two entries in the column coming from the same row, which we assumed doesn’t happen.

Yes, this does change the tableau t, but our conclusion is about the shape of t, which remains the same.

So now we can figure \lambda_1+\dots+\lambda_i as the number of entries in the first i rows of t^\lambda. Since these contain all the entries from the first i rows of s^\mu, it must be greater than or equal to that number. But that number is just as clearly \mu_1+\dots+\mu_i. Since this holds for all i, we conclude that \lambda dominates \mu.

December 20, 2010 Posted by | Algebra, Group theory, Representation Theory, Representations of Symmetric Groups | 1 Comment

The Dominance Order on Partitions

Now we want to introduce a partial order on the collection of partitions called the “dominance order”. Given partitions \lambda=(\lambda_1,\dots,\lambda_k) and \mu=(\mu_1,\dots,\mu_l), we say that \lambda “dominates” \mu — and we write \lambda\trianglerighteq\mu — if


for all i\geq1.

We can interpret this by looking at their Ferrers diagrams. First look at the first rows of the diagrams. Are there more dots in the diagram for \lambda than in that for \mu? If so, so far so good; move on to the second row. Now are there more dots in the first two rows for \lambda than there are in the first two rows for \mu? If so, keep going. If ever there are more dots above the ith row in the Ferrers diagram of \mu than there are in the Ferrers diagram for \lambda, then the condition fails. If ever we run out of rows in one diagram, we just say that any lower rows we need have length zero.

Let’s consider two partitions of 6: \lambda=(4,2) and \mu=(3,2,1). They have the following Ferrers diagrams:


In the first rows, \lambda has four dots while \mu has three, and 4\geq3. Moving on, \lambda has six dots in the first two rows while \mu has five, and 6\geq5. Finally, \lambda has six dots (still!) in the first three rows while \mu also has six, and 6\geq6. Since the inequality holds for all positive i, we find that \lambda dominates \mu.

As another example, let \lambda=(3,3) and \mu=(4,1,1):


This time, the first rows have three and four dots, respectively — \lambda_1\leq\mu_1. We’re on our way to showing that \lambda\trianglelefteq\mu. But the first two rows have six dots and five dots, respectively — \lambda_1+\lambda_2\geq\mu_1+\mu_2. Since the inequality has flipped sides, neither one of these partitions can dominate the other. Dominance is evidently only a partial order, not a total order.

We can put all the partitions of n into a structure called a “Hasse diagram”, which tells us at a glance which partitions dominate which. This is a graph with partitions at the verices. We draw an arrow from \mu to \lambda if \lambda\trianglerighteq\mu. For partitions of 6, this looks like


December 17, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 6 Comments