This is the second set of homework problems Link to main web page for the course


You do not need to hand these problems in, but be prepared to present the solutions in class.

First problem:

Let $G = C_4 = \{ e, g, g^2, g^3 \}$ with $g^4 = e$. Decompose the module $V = {\mathcal L}\{ e - g^2, g - g^3 \}$ into two 1-dimensional submodules.

Shengwei did this in class. She noted that since we are trying to decompose the module into 1-dimensional irreducibles, then the corresponding matrix representation will have the form $$X(g) = \left[ \begin{matrix} a&0\\0&d \end{matrix} \right]$$ where $a, d \in {\mathbb C}$ and since $g^4 = e$, then $X(g^4) = X(e) = X(g)^4$ so $$X(g)^4 = \left[ \begin{matrix} a^4&0\\0&d^4 \end{matrix} \right] = \left[ \begin{matrix} 1&0\\0&1 \end{matrix} \right]$$ So that means that $a \in \{ 1, -1, i, -i \}$. Since an arbitrary element in $V$ is of the form $r(e - g^2) + s(g - g^3)$, then $g (r(e - g^2) + s(g - g^3)) = a (r(e - g^2) + s(g - g^3))$. By comparing coefficients, we see that if $a=1$ or $-1$ then there are are no solutions. If $a=i$ (by choosing $r=1$ arbitrarily), then $r=1, s=i$ produces $$g ((e - g^2) + i(g - g^3)) = i ((e - g^2) + i(g - g^3))$$ is an element that satisfies the equation. If $a=-i$, then $$g ((e - g^2) - i(g - g^3)) = i ((e - g^2) - i(g - g^3))$$ is an element that satisfies the equation. Therefore $$V = {\mathcal L}\{ e - g^2, g - g^3 \} = {\mathcal L}\{ (e - g^2) + i(g - g^3) \} \oplus {\mathcal L}\{ (e - g^2) - i(g - g^3) \}~.$$

Second problem:

Consider the group $S_3$ acting on the polynomial ring ${\mathbb C}[x_1, x_2, x_3]$ making it an infinite dimensional module. An element $f(x_1, x_2, x_3) \in {\mathbb C}[x_1, x_2, x_3]$ is a polynomial in the three variables and the action of $\sigma \in S_3$ is given given by $$\sigma f(x_1, x_2, x_3) = f( x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)} )~.$$ Let ${\mathbb C}[x_1, x_2, x_3]^{S_3}$ be the submodule consisting of $f(x_1, x_2, x_3)$ such that $\sigma f(x_1, x_2, x_3) = f(x_1, x_2, x_3)$ for all $\sigma \in S_3$. Find $dim_q( {\mathbb C}[x_1, x_2, x_3]^{S_3} )$.

So we discussed this problem a lot longer than I expected to in class on May 21 and I only outlined the problem. Step 1: Conjecture the answer. Method 1: figure out the graded dimensions for the first few terms and then look the answer up in the "online integer sequence database." Method 2: recall that a basis for the submodule ${\mathbb C}[x_1, x_2]^{S_2} = $ submodule consisting of those elements invariant under the $S_2$ action is $$\{ x_1^a x_2^b + x_1^b x_2^a\hbox{ for }a \geq b\}~.$$ Use this to guess (big leap here) that a basis for ${\mathbb C}[x_1, x_2, x_3]^{S_3}$ is the set $$\{ R^{S_3}( x_1^a x_2^b x_3^c ) \hbox{ for }a \geq b \geq c \geq 0 \}$$ Step 2: Show that the set above is a basis by demonstrating that it spans and is linear independent. Step 3: Show that generating function for the number of elements in the set $$\{ (a,b,c) : a \geq b \geq c \geq 0 \}$$ is equal to $\left( \frac{1}{(1-q)(1-q^2)} \right) \frac{1}{(1-q^3)}$. Conclusion $$dim_q( {\mathbb C}[x_1, x_2, x_3]^{S_3} ) = \frac{1}{(1-q)(1-q^2)(1-q^3)}$$ On May 26 we will have all the tools necessary to prove this without explicitly producing a basis.