3.12. Lecture 12

3.12.1. Symmetric and alternating products

We defined the tensor product V⊗VV\otimes V as a quotient ℂ⁢(V×V)/U{\mathbb{C}}(V\times V)/U, where the subspace UU is chosen to give the desired linear relations in V⊗VV\otimes V. We will now do something similar to V⊗VV\otimes V, to obtain vector spaces that have even more symmetries.

Definition 3.12.1.

Let U1⊆V⊗VU_{1}\subseteq V\otimes V be the subspace

U1=span⁢{𝐯1⊗𝐯2−𝐯2⊗𝐯1|𝐯1,𝐯2∈V},U_{1}=\mathrm{span}\{{\bf v}_{1}\otimes{\bf v}_{2}-{\bf v}_{2}\otimes{\bf v}_{% 1}\,|\,{\bf v}_{1},{\bf v}_{2}\in V\},

and Sym2⁢(V):=(V⊗V)/U1\mathrm{Sym}^{2}(V):=(V\otimes V)/U_{1}. The vector space Sym2⁢(V)\mathrm{Sym}^{2}(V) is called the (second) symmetric product of VV.

If 𝐯1⊗𝐯2∈V⊗V{\bf v}_{1}\otimes{\bf v}_{2}\in V\otimes V, then

𝐯1⊗𝐯2+U1=𝐯2⊗𝐯1+(𝐯1⊗𝐯2−𝐯2⊗𝐯1)+U1=𝐯2⊗𝐯1+U1.{\bf v}_{1}\otimes{\bf v}_{2}+U_{1}={\bf v}_{2}\otimes{\bf v}_{1}+\big{(}{\bf v% }_{1}\otimes{\bf v}_{2}-{\bf v}_{2}\otimes{\bf v}_{1}\big{)}+U_{1}={\bf v}_{2}% \otimes{\bf v}_{1}+U_{1}. (3.12.2)

Thus, modding out the subspace U1U_{1} enforces the relation “𝐯1⊗𝐯2=𝐯2⊗𝐯1{\bf v}_{1}\otimes{\bf v}_{2}={\bf v}_{2}\otimes{\bf v}_{1}” on V⊗VV\otimes V. We normally write 𝐯1⁢𝐯2{\bf v}_{1}{\bf v}_{2} for the element 𝐯1⊗𝐯2+U1{\bf v}_{1}\otimes{\bf v}_{2}+U_{1} of Sym2⁢(V)\mathrm{Sym}^{2}(V). We have 𝐯1⁢𝐯2=𝐯2⁢𝐯1{\bf v}_{1}{\bf v}_{2}={\bf v}_{2}{\bf v}_{1}, as well as linear relations in both arguments. As a consequence we have that, if 𝐯1,…,𝐯n{\bf v}_{1},\ldots,{\bf v}_{n} is a basis of VV then

{𝐯i⁢𝐯j| 1≤i≤n, 1≤j≤i}\{{\bf v}_{i}{\bf v}_{j}\,|\,1\leq i\leq n,\;1\leq j\leq i\}

is a basis of Sym2⁢(V)\mathrm{Sym}^{2}(V). Thus dim⁡(Sym2⁢(V))=n⁢(n+1)2\operatorname{dim}\big{(}\mathrm{Sym}^{2}(V)\big{)}=\frac{n(n+1)}{2}.

In a similar fashion, we will also define a subspace U2U_{2} of V⊗VV\otimes V that enforces the relation “𝐯1⊗𝐯2=−𝐯2⊗𝐯1{\bf v}_{1}\otimes{\bf v}_{2}=-{\bf v}_{2}\otimes{\bf v}_{1}” on V⊗VV\otimes V:

Definition 3.12.3.

Let U2⊆V⊗VU_{2}\subseteq V\otimes V be the subspace

U2:=span⁢{𝐯1⊗𝐯2+𝐯2⊗𝐯1|𝐯1,𝐯2∈V},U_{2}:=\mathrm{span}\{{\bf v}_{1}\otimes{\bf v}_{2}+{\bf v}_{2}\otimes{\bf v}_% {1}\,|\,{\bf v}_{1},{\bf v}_{2}\in V\},

and ⋀2V:=(V⊗V)/U2\bigwedge^{2}V:=(V\otimes V)/U_{2}. The vector space ⋀2(V)\bigwedge^{2}(V) is called the (second) alternating product of VV.

By the same method as the computation 3.12.2 above, we have 𝐯1⊗𝐯2+U2=−𝐯2⊗𝐯1+U2{\bf v}_{1}\otimes{\bf v}_{2}+U_{2}=-{\bf v}_{2}\otimes{\bf v}_{1}+U_{2}. We normally write 𝐯1∧𝐯2{\bf v}_{1}\wedge{\bf v}_{2} for 𝐯1⊗𝐯2+U2{\bf v}_{1}\otimes{\bf v}_{2}+U_{2}; and 𝐯1∧𝐯2{\bf v}_{1}\wedge{\bf v}_{2} is linear both arguments, and 𝐯1∧𝐯2=−𝐯2∧𝐯1{\bf v}_{1}\wedge{\bf v}_{2}=-{\bf v}_{2}\wedge{\bf v}_{1}. Thus, if we let 𝐯1,…,𝐯n{\bf v}_{1},\ldots,{\bf v}_{n} be a basis of VV, we have

{𝐯i∧𝐯j| 1≤i≤n, 1≤j<i}\{{\bf v}_{i}\wedge{\bf v}_{j}\,|\,1\leq i\leq n,\;1\leq j<i\}

is a basis of ⋀2(V)\bigwedge^{2}(V). Thus dim⁡(⋀2(V))=n⁢(n−1)2\operatorname{dim}\big{(}\bigwedge^{2}(V)\big{)}=\frac{n(n-1)}{2}.

We now look at representations on these spaces:

Proposition 3.12.4.

Let (π,V)(\pi,V) be a representation of a group GG. Then (π⊗π,U1)(\pi\otimes\pi,U_{1}) and (π⊗π,U2)(\pi\otimes\pi,U_{2}) are subrepresentations of (π⊗π,V⊗V)(\pi\otimes\pi,V\otimes V), and

(π⊗π,V⊗V)≅(Sym2⁢π,Sym2⁢(V))⊕(π∧π,⋀2(V)).(\pi\otimes\pi,V\otimes V)\cong\big{(}\mathrm{Sym}^{2}\pi,\mathrm{Sym}^{2}(V)% \big{)}\oplus\big{(}\pi\wedge\pi,\bigwedge\!\!^{2}(V)\big{)}.
Proof.

See Problem 3.12.3 below. ∎

Proposition 3.12.5.

Let (π,V)(\pi,V) be a finite-dimensional representation of a finite group GG. Then for all g∈Gg\in G,

χSym2⁢π⁢(g)=12⁢(χπ⁢(g)2+χπ⁢(g2)),χπ∧π⁢(g)=12⁢(χπ⁢(g)2−χπ⁢(g2)).\chi_{\mathrm{Sym}^{2}\pi}(g)=\frac{1}{2}\big{(}\chi_{\pi}(g)^{2}+\chi_{\pi}(g% ^{2})\big{)},\qquad\chi_{\pi\wedge\pi}(g)=\frac{1}{2}\big{(}\chi_{\pi}(g)^{2}-% \chi_{\pi}(g^{2})\big{)}.
Proof.

We only need to prove the first of these identities, the other follows from Proposition 3.12.4 and Proposition 2.7.9 (iii).

Let 𝐯1,…,𝐯n{\bf v}_{1},\ldots,{\bf v}_{n} be an eigenbasis of VV for π⁢(g)\pi(g), i.e. π⁢(g)⁢𝐯i=λi⁢𝐯i\pi(g){\bf v}_{i}=\lambda_{i}{\bf v}_{i}. Then

{𝐯i⁢𝐯j| 1≤i≤n, 1≤j≤i}\{{\bf v}_{i}{\bf v}_{j}\,|\,1\leq i\leq n,\;1\leq j\leq i\}

is a basis of Sym2⁢(V)\mathrm{Sym}^{2}(V). This is an eigenbasis for Sym2⁢π⁢(g)\mathrm{Sym}^{2}\pi(g) as

Sym2⁢π⁢(g)⁢𝐯i⁢𝐯j=(λi⁢𝐯i)⁢(λj⁢𝐯j)=(λi⁢λj)⁢𝐯i⁢𝐯j.\mathrm{Sym}^{2}\pi(g){\bf v}_{i}{\bf v}_{j}=(\lambda_{i}{\bf v}_{i})(\lambda_% {j}{\bf v}_{j})=(\lambda_{i}\lambda_{j}){\bf v}_{i}{\bf v}_{j}.

The definition of χSym2⁢π⁢(g)\chi_{\mathrm{Sym}^{2}\pi}(g) gives

χSym2⁢π⁢(g)=∑i=1n∑j=1iλi⁢λj.\chi_{\mathrm{Sym}^{2}\pi}(g)=\sum_{i=1}^{n}\sum_{j=1}^{i}\lambda_{i}\lambda_{% j}.

On the other hand,

χπ⁢(g)2=(∑k=1nλk)2\displaystyle\chi_{\pi}(g)^{2}=\left(\sum_{k=1}^{n}\lambda_{k}\right)^{2} =∑i=1nλi2+2⁢∑i=1n∑1≤j<iλi⁢λj\displaystyle=\sum_{i=1}^{n}\lambda_{i}^{2}+2\sum_{i=1}^{n}\sum_{1\leq j<i}% \lambda_{i}\lambda_{j}
=χπ⁢(g2)+2⁢(∑i=1n∑1≤j≤iλi⁢λj−∑i=1nλi2)\displaystyle=\chi_{\pi}(g^{2})+2\left(\sum_{i=1}^{n}\sum_{1\leq j\leq i}% \lambda_{i}\lambda_{j}-\sum_{i=1}^{n}\lambda_{i}^{2}\right)
=2⁢χSym2⁢π⁢(g)−χπ⁢(g2).\displaystyle=2\chi_{\mathrm{Sym}^{2}\pi}(g)-\chi_{\pi}(g^{2}).

∎

We conclude this section by noting that we can generalise these constructions to higher symmetric and alternating powers Symk⁢(V)\mathrm{Sym}^{k}(V), ⋀kV\bigwedge^{k}V for k=3,4,…k=3,4,\ldots.

3.12.2. Example: the character table of S5S_{5}

We will now complete the character table of S5S_{5}. The irreducibles we already know are the trivial representation triv{\mathrm{triv}}, the sign representation sgn\operatorname{sgn}, the irreducible permutation representation (π,W0)(\pi,W_{0}), and its twist (sgn⁡π,W0)(\operatorname{sgn}\pi,W_{0}), as before. So we can put these entries into the table:

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)1101520203024triv1111111sgn1−111−1−11π4201−10−1sgn⁡π4−20110−1\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1&1&1\\ \operatorname{sgn}&1&-1&1&1&-1&-1&1\\ \pi&4&2&0&1&-1&0&-1\\ \operatorname{sgn}\pi&4&-2&0&1&1&0&-1\end{array}

We then try π∧π\pi\wedge\pi, which has character as shown below. This is an irreducible character (as it has norm one).

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)1101520203024π∧π60−20001\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\pi\wedge\pi&6&0&-2&0&0&0&1\end{array}

A natural next step could be to try sgn⁡(π∧π)\operatorname{sgn}(\pi\wedge\pi), but it turns out this is isomorphic to π∧π\pi\wedge\pi. We can also try Sym2⁡π\operatorname{Sym}^{2}\pi, which has character below; it isn’t irreducible.

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)1101520203024Sym2⁡π10421100\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\operatorname{Sym}^{2}\pi&10&4&2&1&1&0&0\end{array}

By taking inner products with the characters we’ve already found, we see that

Sym2⁢π≅triv⊕π⊕ρ,\mathrm{Sym}^{2}\pi\cong{\mathrm{triv}}\oplus\pi\oplus\rho,

where ρ\rho is another irreducible representation. Finally, we get one more from twisting ρ\rho by sgn\operatorname{sgn}.

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)1101520203024ρ511−11−10sgn⁡ρ5−11−1−110\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\rho&5&1&1&-1&1&-1&0\\ \operatorname{sgn}\rho&5&-1&1&-1&-1&1&0\end{array}

This gives all of the irreducible characters, which we assemble into Table 3.1.

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)1101520203024triv1111111sgn1−111−1−11π4201−10−1sgn⁡π4−20110−1π∧π60−20001ρ511−11−10sgn⁡ρ5−11−1−110\begin{array}[]{c|ccccccc|}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1&1&1\\ \operatorname{sgn}&1&-1&1&1&-1&-1&1\\ \pi&4&2&0&1&-1&0&-1\\ \operatorname{sgn}\pi&4&-2&0&1&1&0&-1\\ \pi\wedge\pi&6&0&-2&0&0&0&1\\ \rho&5&1&1&-1&1&-1&0\\ \operatorname{sgn}\rho&5&-1&1&-1&-1&1&0\\ \end{array}
Table 3.1. Character table of S5S_{5}

3.12.3. Exercises

.

Problem 72. Let (π,V)(\pi,V) be a finite-dimensional representation of a finite group GG. Define U1,U2⊆V⊗VU_{1},U_{2}\subseteq V\otimes V by

U1=span⁢{𝐯1⊗𝐯2−𝐯2⊗𝐯1|𝐯1,𝐯2∈V},U2=span⁢{𝐯1⊗𝐯2+𝐯2⊗𝐯1|𝐯1,𝐯2∈V}U_{1}=\mathrm{span}\{{\bf v}_{1}\otimes{\bf v}_{2}-{\bf v}_{2}\otimes{\bf v}_{% 1}\,|\,{\bf v}_{1},{\bf v}_{2}\in V\},\quad U_{2}=\mathrm{span}\{{\bf v}_{1}% \otimes{\bf v}_{2}+{\bf v}_{2}\otimes{\bf v}_{1}\,|\,{\bf v}_{1},{\bf v}_{2}% \in V\}

Show that (π⊗π,U1)(\pi\otimes\pi,U_{1}) and (π⊗π,U2)(\pi\otimes\pi,U_{2}) are subrepresentations of (π⊗π,V⊗V)(\pi\otimes\pi,V\otimes V) and thus show that

(π⊗π,V⊗V)≅(Sym2⁢π,Sym2⁢(V))⊕(π∧π,⋀2V).(\pi\otimes\pi,V\otimes V)\cong\big{(}\mathrm{Sym}^{2}\pi,\mathrm{Sym}^{2}(V)% \big{)}\oplus\big{(}\pi\wedge\pi,\textstyle\bigwedge^{2}V\big{)}.

.

Problem 73. Let (π,ℂ2)(\pi,{\mathbb{C}}^{2}) be a representation of a group GG. Suppose that the matrix of π⁢(g)\pi(g) with respect to the basis {𝐛1,𝐛2}\{{\bf b}_{1},{\bf b}_{2}\} is given by

π⁢(g)=(abcd).\pi(g)=\left(\begin{matrix}a&b\\ c&d\end{matrix}\right).
  1. (a)
    ​

    Compute the matrix of Sym2⁢π⁢(g)\mathrm{Sym}^{2}\pi(g) with respect to the basis {𝐛12,𝐛1⁢𝐛2,𝐛22}\{{\bf b}_{1}^{2},{\bf b}_{1}{\bf b}_{2},{\bf b}_{2}^{2}\}.

  2. (b)
    ​

    Compute the matrix of π∧π⁢(g)\pi\wedge\pi(g) with respect to the basis {𝐛1∧𝐛2}\{{\bf b}_{1}\wedge{\bf b}_{2}\}.

Problem 74. Let (σ,V)(\sigma,V) be any irreducible 5-dimensional representation of S5S_{5}. Decompose (Sym2⁡σ,Sym2⁡V)(\operatorname{Sym}^{2}\sigma,\operatorname{Sym}^{2}V) and (σ∧σ,Λ2⁢V)(\sigma\wedge\sigma,\Lambda^{2}V) into irreducible representations.

Problem 75. Let (π,ℂ4)(\pi,{\mathbb{C}}^{4}) be the permutation representation of S4S_{4}, and let (ρ,V)(\rho,V) be the 2-dimensional irreducible representation of S4S_{4}.

  1. (a)
    ​

    Show that (Sym2⁡π,Sym2⁡ℂ4)(\operatorname{Sym}^{2}\pi,\operatorname{Sym}^{2}{\mathbb{C}}^{4}) has a unique subrepresentation isomorphic to (ρ,V)(\rho,V).

  2. (b)
    ​

    Use the ρ\rho-isotopic projector to find that subrepresentation.

Problem 76. A group GG of order 168168 has conjugacy classes C1C_{1}, C2C_{2}, C3C_{3}, C4C_{4}, C7⁢AC_{7A} and C7⁢BC_{7B}, where each conjugacy class is labelled by the order of any element in that class (so, for example, any element of C7⁢AC_{7A} or C7⁢BC_{7B} has order 77). The following shows one of the rows of the character table of GG.

size:12156422424classC1C2C3C4C7⁢AC7⁢B(π1,V)3−101−1+−72−1−−72\begin{array}[]{c|cccccc}\text{size:}&1&21&56&42&24&24\\ \text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr(\pi_{1},V)&3&-1&0&1&\frac{-1+\sqrt{-7}}{2}&\frac{-1-\sqrt{-7}}{2}\\ \end{array}
  1. (a)
    ​

    Show that if xx is an element of C7⁢AC_{7A} or C7⁢BC_{7B}, then xx is conjugate to x2x^{2}.

  2. (b)
    ​

    Find the character table of GG.