# Young projectors in Fulton and Harris

In Section 4 of Fulton and Harris' book Representation Theory, they give the definition of a Young tableau of shape $\lambda = (\lambda_1,\dots,\lambda_k)$ and then define two subgroups of $S_d$, the symmetric group on $d$ letters: $$P = P_\lambda = \{g\in S_d : g\ \text{preserves each row}\}$$ and $$Q = Q_\lambda = \{g\in S_d : g\ \text{preserves each column}\}.$$ In the group algebra $\Bbb CS_d$, they define two elements corresponding to these subgroups: $$a_\lambda = \sum_{g\in P}e_g \quad\text{and}\quad b_\lambda = \sum_{g\in Q}\operatorname{sgn}(g)\cdot e_g.$$ They say

To see what $a_\lambda$ and $b_\lambda$ do, observe that if $V$ is any vector space and $S_d$ acts on the $d$th tensor power $V^{\otimes d}$ by permuting factors, the image of the element $a_\lambda\in\Bbb CS_d\to \operatorname{End}(V^{\otimes d})$ is just the subspace $$\operatorname{Im}(a_\lambda) = \operatorname{Sym}^{\lambda_1}V\otimes\dotsb\otimes\operatorname{Sym}^{\lambda_k}V\subset V^{\otimes d},$$ where the inclusion on the right is obtained by grouping the factors of $V^{\otimes d}$ according to the rows of the Young tableaux. Similarly, the image of $b_\lambda$ on this tensor power is $$\operatorname{Im}(b_\lambda) = \wedge{}^{\mu_1}V\otimes\dotsb\otimes\wedge{}^{\mu_l}V\subset V^{\otimes d},$$ where $\mu = \left(\mu_1, \ldots, \mu_l\right)$ is the conjugate partition to $\lambda$.

I am very confused.

1. What is the meaning of the notation $a_\lambda\in \Bbb CS_d\to\operatorname{End}(V^{\otimes d})$? I was operating under the assumption that this was just a typo or sloppy notation. To my understanding, $a_\lambda$ can be thought of as an endomorphism of the $d$th tensor power by naturally permuting the coordinates of the tensors it sees, but this looks like it is saying that $a_\lambda$ is a map $\Bbb CS_d\to\operatorname{End}(V^{\otimes d})$, which isn't the same thing, as far as I can tell.

2. How do we see that $$\operatorname{Im}(a_\lambda) = \operatorname{Sym}^{\lambda_1}V\otimes\dotsb\otimes\operatorname{Sym}^{\lambda_k}V\subset V^{\otimes d}?$$ I worked this out in a really simple case with $\lambda = (2,2)$ and numbering the Young diagram 1 to 4, but this didn't shed any insight on the problem. I guess I am probably confused about what this is even saying; to my knowledge this statement is saying that anything in the image of $a_\lambda$ has the property that if I hit it with a permutation $\sigma\in S_{\lambda_i+1,\dots,\lambda_{i+1}}$ (that is, a permutation that only permutes components $\lambda_i+1$ through $\lambda_{i+1}$ in the $d$th tensor power), then it is unchanged. A proof of this (or whatever the correct interpretation is) would be much appreciated.

3. The same question as 2., except for $$\operatorname{Im}(b_\lambda) = \wedge{}^{\mu_1}V\otimes\dotsb\otimes\wedge{}^{\mu_l}V\subset V^{\otimes d}.$$

Don't forget to say what you mean by "each row" and "each column". I assume you are referring to some fixed tableau of shape $\lambda$ here.

I shall use the notation $\lambda_{\leq j}$ for the integer $\lambda _{1}+\lambda_{2}+\cdots+\lambda_{j}$, where $j$ is any element of $\left\{ 0,1,\ldots,k\right\}$. Note that $\lambda_{\leq0}=0$ and $\lambda_{\leq k}=n$.

1. Yes, your understanding is right. When Fulton/Harris say "the element $a_{\lambda}\in\mathbb{C}S_{d}\rightarrow\operatorname*{End}\left( V^{\otimes d}\right)$", they mean "the image of the element $a_{\lambda}\in \mathbb{C}S_{d}$ under the canonical homomorphism $\mathbb{C}S_{d} \rightarrow\operatorname*{End}\left( V^{\otimes d}\right)$". This image is an endomorphism of $V^{\otimes d}$, and so itself has an image (of course, this is a different meaning of "image"); the latter image is what they call $\operatorname*{Im}\left( a_{\lambda}\right)$. This is an abuse of notation which is not too bad since this is the only reasonable meaning of $\operatorname*{Im}\left( a_{\lambda}\right)$ around (but be careful -- once you have more than one representation of $S_{d}$ around, you're screwed).

2. Here is what they mean: For any $m\in\mathbb{N}$, there is a canonical embedding \begin{align*} \iota_{\operatorname*{Sym},m}:\operatorname*{Sym}\nolimits^{m}V & \rightarrow V^{\otimes m},\\ v_{1}v_{2}\cdots v_{m} & \mapsto\sum_{\sigma\in S_{m}}v_{\sigma\left( 1\right) }\otimes v_{\sigma\left( 2\right) }\otimes\cdots\otimes v_{\sigma\left( m\right) } \end{align*} (where the "$v_{1}v_{2}\cdots v_{m}$" on the left hand side means the projection of the tensor $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{m}\in V^{\otimes m}$ onto $\operatorname*{Sym}\nolimits^{m}V$). Thus, you have canonical embeddings $\iota_{\operatorname*{Sym},\lambda_{i}} :\operatorname*{Sym}\nolimits^{\lambda_{i}}V\rightarrow V^{\otimes\lambda_{i} }$ for all $i\in\left\{ 1,2,\ldots,k\right\}$. The tensor product of these $k$ embeddings is an embedding \begin{align*} \iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota_{\operatorname*{Sym} ,\lambda_{2}}\otimes\cdots\otimes\iota_{\operatorname*{Sym},\lambda_{k}} & :\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym} \nolimits^{\lambda_{2}}V\otimes\cdots\otimes\operatorname*{Sym} \nolimits^{\lambda_{k}}V\\ & \rightarrow V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}} \otimes\cdots\otimes V^{\otimes\lambda_{k}}. \end{align*} If we identify $V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}} \otimes\cdots\otimes V^{\otimes\lambda_{k}}$ with $V^{\otimes n}$ in the standard way (i.e., identifying each tensor \begin{align*} & \left( a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\lambda_{1}}\right) \otimes\left( b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\lambda_{2}}\right) \otimes\cdots\otimes\left( g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\lambda_{k} }\right) \\ & \in V^{\otimes\lambda_{1}}\otimes V^{\otimes\lambda_{2}}\otimes\cdots\otimes V^{\otimes\lambda_{k}} \end{align*} (where the letters "$a$" and "$b$" here have nothing to do with $a_{\lambda}$ and $b_{\lambda}$) with \begin{align*} & a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\lambda_{1}}\otimes b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\lambda_{2}}\otimes\cdots\otimes g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\lambda_{k}}\\ & \in V^{\otimes n} \end{align*} ), then this becomes an embedding $$\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym} \nolimits^{\lambda_{2}}V\otimes\cdots\otimes\operatorname*{Sym} \nolimits^{\lambda_{k}}V\rightarrow V^{\otimes n}.$$ Fulton/Harris regard this embedding as an inclusion (another abuse of notation), i.e., they use it to pretend that $\operatorname*{Sym} \nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}\nolimits^{\lambda_{2} }V\otimes\cdots\otimes\operatorname*{Sym}\nolimits^{\lambda_{k}}V$ is a $\mathbb{C}$-vector subspace of $V^{\otimes n}$. Now they claim that the image $\operatorname{Im}\left( a_{\lambda}\right)$ is precisely this subspace. This yields, in particular, your claim that anything in this image is unchanged when you hit it with a permutation $\sigma\in S_{\left\{ \lambda_{\leq i}+1,\ldots,\lambda_{\leq i+1}\right\} }$; but it is a stronger statement.

Why is it true? Well, it isn't true in general. You need to assume that the tableau you are using to define $a_{\lambda}$ is the one whose first row has entries $1,2,\ldots,\lambda_{1}$ (in this order), whose second row has entries $\lambda_{1}+1,\lambda_{1}+2,\ldots,\lambda_{1}+\lambda_{2}$ (in this order), and so on (i.e., if you read it row by row from top to bottom, then you get the sequence $\left( 1,2,\ldots,n\right)$). Then, it is easy to see that the permutations $g\in P_{\lambda}$ are exactly the permutations in $S_{n}$ that can be written in the form $\sigma_{1}\sigma_{2}\cdots\sigma_{k}$, where each $\sigma_{i}$ belongs to $S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }$. Moreover, they can be written in this form uniquely; thus, $P_{\lambda}$ is the following internal direct product of subgroups of $S_{n}$: $$P_{\lambda}=\prod_{i=1}^{k}S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }.$$ Hence, every $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\in V^{\otimes n}$ satisfies \begin{align*} & \sum_{g\in P_{\lambda}}g\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) \\ & =\sum_{\substack{\left( \sigma_{1},\sigma_{2},\ldots,\sigma_{k}\right) ;\\\text{each }\sigma_{i}\text{ belongs to }S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }} }\underbrace{\left( \sigma_{1}\sigma_{2}\cdots\sigma_{k}\right) \left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) }_{\substack{=\bigotimes _{i=1}^{k}\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) \\\text{(where the notation }\bigotimes_{i=1}^{k}w_{i}\text{ is shorthand for }w_{1}\otimes w_{2}\otimes\cdots\otimes w_{k}\text{,}\\\text{whenever }w_{1}\in V^{\otimes\lambda_{1}},w_{2}\in V^{\otimes\lambda_{2}},\ldots,w_{k}\in V^{\otimes\lambda_{k}}\text{)}}}\\ & =\sum_{\substack{\left( \sigma_{1},\sigma_{2},\ldots,\sigma_{k}\right) ;\\\text{each }\sigma_{i}\text{ belongs to }S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} }}}\bigotimes _{i=1}^{k}\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) \\ & =\bigotimes_{i=1}^{k}\underbrace{\sum_{\sigma_{i}\in S_{\left\{ \lambda_{\leq i-1}+1,\lambda_{\leq i-1}+2,\ldots,\lambda_{\leq i}\right\} } }\sigma_{i}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1} +2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) }_{=\iota _{\operatorname*{Sym},\lambda_{i}}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) }\\ & =\bigotimes_{i=1}^{k}\iota_{\operatorname*{Sym},\lambda_{i}}\left( v_{\lambda_{\leq i-1}+1}\otimes v_{\lambda_{\leq i-1}+2}\otimes\cdots\otimes v_{\lambda_{\leq i}}\right) \\ & =\left( \iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota _{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota _{\operatorname*{Sym},\lambda_{k}}\right) \left( v_{1}\otimes v_{2} \otimes\cdots\otimes v_{n}\right) . \end{align*} Since $\sum_{g\in P_{\lambda}}g=a_{\lambda}$, this rewrites as $$a_{\lambda}\left( v_{1}\otimes v_{2}\otimes\cdots\otimes v_{n}\right) =\left( \iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota _{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota _{\operatorname*{Sym},\lambda_{k}}\right) \left( v_{1}\otimes v_{2} \otimes\cdots\otimes v_{n}\right) .$$ Thus, the image $\operatorname*{Im}\left( a_{\lambda}\right)$ is the image of the map $\iota_{\operatorname*{Sym},\lambda_{1}}\otimes\iota _{\operatorname*{Sym},\lambda_{2}}\otimes\cdots\otimes\iota _{\operatorname*{Sym},\lambda_{k}}$. Since we have agreed to identify the latter image with $\operatorname*{Sym}\nolimits^{\lambda_{1}}V\otimes \operatorname*{Sym}\nolimits^{\lambda_{2}}V\otimes\cdots\otimes \operatorname*{Sym}\nolimits^{\lambda_{k}}V$, we thus conclude that $$\operatorname{Im}\left( a_{\lambda}\right) =\operatorname*{Sym} \nolimits^{\lambda_{1}}V\otimes\operatorname*{Sym}\nolimits^{\lambda_{2} }V\otimes\cdots\otimes\operatorname*{Sym}\nolimits^{\lambda_{k}}V.$$

3. This is similar to part 2. Instead of the embeddings $\iota _{\operatorname*{Sym},m}$, you now need the embeddings \begin{align*} \iota_{\Lambda,m}:\Lambda^{m}V & \rightarrow V^{\otimes m},\\ v_{1}\wedge v_{2}\wedge\cdots\wedge v_{m} & \mapsto\sum_{\sigma\in S_{m} }\left( -1\right) ^{\sigma}v_{\sigma\left( 1\right) }\otimes v_{\sigma\left( 2\right) }\otimes\cdots\otimes v_{\sigma\left( m\right) } \end{align*} (where $\left( -1\right) ^{\sigma}$ is the sign of $\sigma$). These give you canonical embeddings $\iota_{\Lambda,\mu_{i}}:\Lambda^{\mu_{i}}V\rightarrow V^{\otimes\mu_{i}}$ for all $i\in\left\{ 1,2,\ldots,l\right\}$. The tensor product of these $l$ embeddings is an embedding \begin{align*} \iota_{\Lambda,\mu_{1}}\otimes\iota_{\Lambda,\mu_{2}}\otimes\cdots\otimes \iota_{\Lambda,\mu_{l}} & :\Lambda^{\mu_{1}}V\otimes\Lambda^{\mu_{2}} V\otimes\cdots\otimes\Lambda^{\mu_{l}}V\\ & \rightarrow V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes V^{\otimes\mu_{l}}. \end{align*} Identifying $V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes V^{\otimes\mu_{l}}$ with $V^{\otimes n}$ in the standard way, this becomes an embedding $$\Lambda^{\mu_{1}}V\otimes\Lambda^{\mu_{2}}V\otimes\cdots\otimes\Lambda ^{\mu_{l}}V\rightarrow V^{\otimes n}.$$ Fulton/Harris claim that the image $\operatorname{Im}\left( b_{\lambda }\right)$ is precisely the image of this embedding.

This is, again, not true in general. This time, you need to assume that the tableau you are using to define $b_{\lambda}$ is the one whose first column has entries $1,2,\ldots,\mu_{1}$ (in this order), whose second column has entries $\mu_{1}+1,\mu_{1}+2,\ldots,\mu_{1}+\mu_{2}$ (in this order), and so on (i.e., if you read it column by column from left to right, then you get the sequence $\left( 1,2,\ldots,n\right)$). The proof of this is similar to the proof of 2.

This means that the claim in 2. and the claim 3. cannot hold at the same time: The tableau that is required for 2. and the tableau that is required for 3. are different (unless $\lambda$ consists merely of a single row or of a single column). My impression is that Fulton/Harris fix this problem by identifying $V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes V^{\otimes\mu_{l}}$ with $V^{\otimes n}$ in the non-standard way: instead of just "dropping the parentheses", they permute the tensor factors so that the first tensor factors in each parentheses are read first, then the second tensor factors, etc. In other words, they identify each tensor \begin{align*} & \left( a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\mu_{1}}\right) \otimes\left( b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\mu_{2}}\right) \otimes\cdots\otimes\left( g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\mu_{l} }\right) \\ & \in V^{\otimes\mu_{1}}\otimes V^{\otimes\mu_{2}}\otimes\cdots\otimes V^{\otimes\mu_{l}} \end{align*} (where the letters "$a$" and "$b$" here have nothing to do with $a_{\lambda}$ and $b_{\lambda}$) with \begin{align*} & \underbrace{a_{1}\otimes b_{1}\otimes\cdots}_{\text{all tensor factors with subscript }1}\otimes\underbrace{a_{2}\otimes b_{2}\otimes\cdots}_{\text{all tensor factors with subscript }2}\otimes\cdots\otimes\underbrace{a_{\mu_{1} }\otimes b_{\mu_{1}}\otimes\cdots}_{\text{all tensor factors with subscript }\mu_{1}}\\ & \in V^{\otimes n}, \end{align*} instead of with the tensor \begin{align*} & a_{1}\otimes a_{2}\otimes\cdots\otimes a_{\mu_{1}}\otimes b_{1}\otimes b_{2}\otimes\cdots\otimes b_{\mu_{2}}\otimes\cdots\otimes g_{1}\otimes g_{2}\otimes\cdots\otimes g_{\mu_{l}}\\ & \in V^{\otimes n} \end{align*} as we did above. If you make this change, then the tableau that is required for 3. becomes the same as that required for 2.

4. Get a better text. Fulton/Harris are known for handwaving and half-truths. Fulton's Young tableaux is much more readable everywhere I've tried reading it, although I don't know if it covers the stuff you need.

• Wow. Really awesome answer. Thanks so much for putting in all the effort to write this up. I think that one of the main points I was stuck at was not knowing (or having forgotten) about these canonical embeddings you mentioned. I'll try and find a copy of Fulton's book. Do you have any recommendations for books on representation theory? I'm mostly interested in representations of $S_n$ and then some stuff on representations of Lie groups. – Alex Ortiz May 17 '18 at 23:38
• @AOrtiz: I'm not an expert on representation theory myself, so don't take my advice for too much. I have read Etingof et al. and liked it (but it is terse; I had to ask the author on occasion...). There are various other texts these days (Steinberg, Procesi), and lecture notes (Snowden, Savage). – darij grinberg May 17 '18 at 23:40