Brian Bi
$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})}$

Problem 3.9.1

1. The map $$\rho_U : A \to \End_k U$$ is obviously linear whenever $$f$$ is. In order for $$\rho_U$$ to be an algebra homomorphism, it is sufficient and necessary that, additionally, $$\rho_U(ab) = \rho_U(a)\rho_U(b)$$ for all $$a, b \in A$$. That is $\begin{pmatrix} \rho_V(ab) & f(ab) \\ 0 & \rho_W(ab) \end{pmatrix} = \begin{pmatrix} \rho_V(a) & f(a) \\ 0 & \rho_W(a) \end{pmatrix} \begin{pmatrix} \rho_V(b) & f(b) \\ 0 & \rho_W(b) \end{pmatrix}$ Using the fact that $$\rho_V(ab) = \rho_V(a)\rho_V(b)$$ and $$\rho_W(ab) = \rho_W(a)\rho_W(b)$$, and expanding out the RHS, we arrive at the equivalent condition $f(ab) = \rho_V(a)f(b) + f(a)\rho_W(b)$ We make one useful observation here before proceeding. Substituting $$a = b = 1$$ yields $$f(1) = \Id_V f(1) + f(1) \Id_W = 2f(1)$$, therefore $$f(1) = 0$$.

2. For all $$a, b \in A$$, we have \begin{align*} \mathrm{d}X(ab) &= \rho_V(ab)X - X\rho_W(ab) \\ &= \rho_V(a)\rho_V(b)X - X\rho_W(a)\rho_W(b) \\ &= \rho_V(a)\rho_V(b)X - \rho_V(a)X\rho_W(b) + \rho_V(a)X\rho_W(b) - X\rho_W(a)\rho_W(b) \\ &= \rho_V(a)(\rho_V(b)X - X\rho_W(b)) + (\rho_V(a)X - X\rho_W(a))\rho_W(b) \\ &= \rho_V(a)\mathrm{d}X(b) + \mathrm{d}X(a)\rho_W(b) \end{align*} Therefore $$\mathrm{d}X$$ is a cocycle.

The map $$\mathrm{d}X$$ will be identically zero iff for all $$a \in A$$, $$w \in W$$, we have $$\rho_V(a)Xw = X\rho_W(a)w$$. This is the same as the definition of $$X$$ being a homomorphism between the representations $$\rho_V$$ and $$\rho_W$$.

The map $$\mathrm{d} : \Hom_k(W, V) \to \Hom_k(A, \Hom_k(W, V))$$ is easily seen to be linear. We have determined that $$\ker \mathrm{d} = \Hom_A(W, V)$$. The first isomorphism theorem implies that $$\im \mathrm{d} \cong \Hom_k(W, V)/\Hom_A(W, V)$$.

3. Suppose $$f, f' \in Z^1(W, V)$$ with $$f - f' \in B^1(W, V)$$. The corresponding representations are $\rho_U(a) = \begin{pmatrix} \rho_V(a) & f(a) \\ 0 & \rho_W(a) \end{pmatrix} \qquad \rho_{U'}(a) = \begin{pmatrix} \rho_V(a) & f'(a) \\ 0 & \rho_W(a) \end{pmatrix}$ Suppose $$X \in \Hom_k(W, V)$$. Define $\varphi_X = \begin{pmatrix} \Id_V & X \\ 0 & \Id_W \end{pmatrix}$ $$\varphi_X$$ is always invertible, therefore it is always an isomorphism of $$U$$ and $$U'$$ as vector spaces. For each $$a \in A$$, \begin{align*} \rho_U(a) \varphi_X - \varphi_X \rho_{U'}(a) &= \begin{pmatrix} \rho_V(a) & f(a) \\ 0 & \rho_W(a) \end{pmatrix} \begin{pmatrix} \Id_V & X \\ 0 & \Id_W \end{pmatrix} - \begin{pmatrix} \Id_V & X \\ 0 & \Id_W \end{pmatrix} \begin{pmatrix} \rho_V(a) & f'(a) \\ 0 & \rho_W(a) \end{pmatrix} \\ &= \begin{pmatrix} \rho_V & \rho_V(a) X + f(a) \\ 0 & \rho_W \\ \end{pmatrix} - \begin{pmatrix} \rho_V & f'(a) + X\rho_W(a) \\ 0 & \rho_W \\ \end{pmatrix} \\ &= \begin{pmatrix} 0 & (\rho_V(a) X - X \rho_W(a)) - (f'(a) - f(a)) \\ 0 & 0 \\ \end{pmatrix} \\ &= 0 \end{align*} therefore $$\varphi_X : U \to U'$$ is an isomorphism of representations if and only if $$\rho_V(a)X - X\rho_W(a) = f'(a) - f(a)$$ for all $$a \in A$$, or in other words if and only if $$f' - f = \mathrm{d}X$$.

Therefore, if $$f' - f \in B^1(W, V)$$, then with the choice of $$X$$ such that $$\mathrm{d}X = f' - f$$, we obtain the isomorphism of representations $$\varphi_X : U \to U'$$; conversely, if $$U$$ and $$U'$$ are isomorphic representations with the isomorphism of the form $$\varphi_X$$ for some $$X$$ then it must be that $$f' - f = \mathrm{d}X$$, so that $$f' - f \in B^1(W, V)$$.

4. Suppose $$f, f' \in \mathrm{Ext}^1(W, V)$$ with $$f' = \lambda f$$ where $$\lambda \in k^\times$$. We can choose representatives $$g, g' \in Z^1(W, V)$$ of $$f, f'$$ respectively so that $$g' = \lambda g$$ also. Define $$\varphi : U_g \to U_{g'}$$ by $\varphi = \begin{pmatrix} \Id_V & 0 \\ 0 & \lambda \Id_W \end{pmatrix}$ It is easily verified that for each $$a \in A$$, $$\varphi \circ \rho_{U_g}(a) = \rho_{U_{g'}}(a) \circ \varphi$$, so $$\varphi$$ is a homomorphism of representations. The matrix of $$\varphi$$ is invertible, so the representations $$U_g, U_{g'}$$ are isomorphic.

Conversely, suppose $$g, g' \in Z^1(W, V)$$ are given and that there is an isomorphism $$\varphi : U_g \to U_{g'}$$. In block form, suppose $\varphi = \begin{pmatrix} \varphi_1 & \varphi_2 \\ \varphi_3 & \varphi_4 \end{pmatrix}$ The condition $$\varphi \circ U_g(a) = U_{g'}(a) \circ \varphi$$ for all $$a \in A$$ implies, after expanding both sides and equating blocks on each side, \begin{align} \varphi_1 \rho_V(a) &= \rho_V(a) \varphi_1 + g'(a) \varphi_3 \label{eqn:iso1} \\ \varphi_1 g(a) + \varphi_2 \rho_W(a) &= \rho_V(a) \varphi_2 + g'(a) \varphi_4 \label{eqn:iso2} \\ \varphi_3 \rho_V(a) &= \rho_W(a) \varphi_3 \label{eqn:iso3} \\ \varphi_3 g(a) + \varphi_4 \rho_W(a) &= \rho_W(a) \varphi_4 \label{eqn:iso4} \end{align} for all $$a \in A$$.

$$\pref{eqn:iso3}$$ implies that $$\varphi_3 : V \to W$$ is a homomorphism of representations. By Schur's lemma, either $$\varphi_3 = 0$$, or $$\varphi_3$$ is an isomorphism.

Consider first the case that $$\varphi_3 = 0$$. Then $$\pref{eqn:iso1}$$ and $$\pref{eqn:iso4}$$ respectively simplify to \begin{align*} \varphi_1 \rho_V(a) &= \rho_V(a) \varphi_1 \\ \varphi_4 \rho_W(a) &= \rho_W(a) \varphi_4 \end{align*} so $$\varphi_1$$ and $$\varphi_4$$ are respectively endomorphisms of the representations $$V$$ and $$W$$. By Schur's lemma, each is a scalar operator, say, $$\varphi_1 = \lambda \Id_V$$ and $$\varphi_4 = \mu \Id_W$$. Then $$\pref{eqn:iso2}$$ simplifies to $\lambda g(a) + \varphi_2 \rho_W(a) = \rho_V(a) \varphi_2 + \mu g'(a)$ that is $\rho_V(a) \varphi_2 - \varphi_2 \rho_W(a) = \mathrm{d}\varphi_2 = \lambda g(a) - \mu g'(a)$ so $$\lambda g - \mu g' \in B^1(W, V)$$. Let $$f, f'$$ denote the projections of $$g, g'$$ onto $$\mathrm{Ext}^1(W, V)$$; then $$f' = \frac{\lambda}{\mu}f$$.

(Travis Hance supplied this part of the proof.) On the other hand, if $$\varphi_3$$ is an isomorphism, then left-multiply both sides of $$\pref{eqn:iso4}$$ by $$\varphi_3^{-1}$$ to give $g(a) + \varphi_3^{-1} \varphi_4 \rho_W(a) = \varphi_3^{-1}\rho_W(a)\varphi_4$ Now $$\varphi^{-1}$$ is an isomorphism from $$W$$ to $$V$$, so $$\varphi_3^{-1}\rho_W(a) = \rho_V(a) \varphi_3^{-1}$$. Consequently $g(a) = \varphi_3^{-1} \rho_W(a) \varphi_4 - \varphi_3^{-1} \varphi_4 \rho_W(a) = \rho_V(a) \varphi_3^{-1}\varphi_4 - \varphi_3^{-1}\varphi_4 \rho_W(a) = \mathrm{d}(\varphi_3^{-1}\varphi_4)$ so $$g \in B^1(W, V)$$. Similarly, $$\pref{eqn:iso1}$$ will show that $$g' \in B^1(W, V)$$. Therefore $$f = f' = 0$$.

Problem 3.9.2

1. Suppose $$f$$ is a cocycle, $$f(PQ) = \rho_b(P)f(Q) + f(P)\rho_a(Q) = P(b)f(Q) + Q(a)f(P) \label{eqn:coc}$$ where we have written $$\rho_a, \rho_b$$ for $$\rho_{V_a}, \rho_{V_b}$$ for brevity. To proceed, we analyze two cases.

First, suppose that there exists some $$i$$ with $$a_i \ne b_i$$. Without loss of generality, say $$a_1 \ne b_1$$. Let $$j \in \{1, \ldots, n\}$$ be given. Then $$\pref{eqn:coc}$$ implies \begin{align*} f(x_1 x_j) &= b_1 f(x_j) + a_j f(x_1) \\ f(x_j x_1) &= b_j f(x_1) + a_1 f(x_j) \end{align*} Since $$x_1 x_j = x_j x_1$$, rearranging yields $f(x_j) = (b_j - a_j)\frac{f(x_1)}{b_1 - a_1}$ Let $$\lambda = f(x_1)/(b_1 - a_1)$$. It follows that for all $$j \in \{1, \ldots, n\}$$, we have $f(x_j) = (b_j - a_j)\lambda$ The choice of the map $$\lambda : W \to V$$ therefore suffices to determine the function $$f : A \to \Hom_{\mathbb{C}}(W, V)$$. Define $$f' : A \to \Hom_{\mathbb{C}}(V_a, V_b)$$ by $f'(P) = (P(b) - P(a))\lambda$ Then $$f'$$ is linear in its argument $$P$$, agrees with $$f$$ when $$P = x_i$$ for some $$i$$, and satisfies $$\pref{eqn:coc}$$, so $$f = f'$$. Since $$\Hom_{\mathbb{C}}(V_a, V_b)$$ from which $$\lambda$$ is drawn is one-dimensional, $$Z^1(V_a, V_b)$$ is also one-dimensional. But observe that $$\mathrm{d}\lambda(P) = \rho_b(P) \lambda - \lambda \rho_a(P) = (P(b) - P(a))\lambda = f$$, so every such $$f$$ is a coboundary. Therefore $$\mathrm{Ext}^1(V_a, V_b)$$ is trivial.

On the other hand, if $$a_i = b_i$$ for all $$i$$, then in fact the $$n$$ operators $$\lambda_i = f(x_i)$$, which determine $$f$$, can be chosen independently. The $$f$$ that satisfies $$\pref{eqn:coc}$$ and $$f(x_i) = \lambda_i$$ for each $$i$$ is given by $f(P) = \sum_{i=1}^n \lambda_i P_i(a)$ where $$P_i$$ denotes the (formal) partial derivative $$\d P/\d x_i$$. The space $$Z^1(V_a, V_a)$$ therefore has dimension $$n$$. It is easy to check that $$B^1(V_a, V_a) = 0$$ (indeed, $$B^1(V, V)$$ must be trivial for any one-dimensional representation of any algebra), so $$\mathrm{Ext}^1(V_a, V_a)$$ has dimension $$n$$. Extensions of $$V_a$$ by $$V_a$$ are therefore parametrized by $$[c_1 : \ldots : c_n] \in \mathbb{P}\mathbb{C}^n$$ with $f(P) = \sum_{i=1}^n c_i P_i(a) \Id$

With these results we can classify the two-dimensional representations of $$A$$. Since $$A$$ is commutative, only the one-dimensional representations are irreducible. Let $$V$$ be a two-dimensional representation, let $$V_b$$ be a one-dimensional subrepresentation, and let $$V_a = V/V_b$$. If $$a \ne b$$ then $$\mathrm{Ext}^1(V_a, V_b) = 0$$ so the extension $$V$$ is trivial, that is, $$V \cong V_a \oplus V_b$$. In the case where $$a = b$$, we likewise have the representations $$V_a \oplus V_a$$. Write $$V_{ab} = V_a \oplus V_b$$, where $$a, b$$ may be the same or different.

Since $$\mathrm{Ext}^1(V_a, V_a)$$ is nontrivial, in the case $$a = b$$ we also have indecomposable representations of the form $\rho_V(x_i) = \begin{pmatrix} a_i & c_i \\ 0 & a_i \end{pmatrix}$ where $$[c_1 : \ldots : c_n] \in \mathbb{P}\mathbb{C}^n$$. Denote such a representation by $$U_{ac}$$.

We have therefore two kinds of two-dimensional representations,

1. The representations $$V_{ab} = V_a \oplus V_b$$ where $$a, b \in \mathbb{C}^n$$
2. The representations $$U_{ac}$$ where $$a \in \mathbb{C}^n, b \in \mathbb{P}\mathbb{C}^n \setminus \{0\}$$

No $$V_{ab}$$ is isomorphic to any $$U_{a'c}$$ since the latter are all indecomposable while the former are not. The Krull–Schmidt theorem implies that $$V_{ab} \cong V_{a'b'}$$ if and only if $$a, b$$ is a permutation of $$a', b'$$. The result of Problem 3.9.1(d) implies that $$U_{ac}$$ and $$U_{ac'}$$ are isomorphic only when $$c = c'$$. Finally, when $$a \ne a'$$, the representations $$U_{ac}$$ and $$U_{a'c'}$$ cannot be isomorphic since if $$a_i \ne a'_i$$ then $$x_i$$ will have a different eigenvalue in the two representations.

2. The algebra $$B$$ has a unique irreducible representation $$V$$, in which $$\rho_V(x_i) = 0$$ for all $$i$$. Suppose $$f$$ is a cocycle. Because $$x_i x_j = 0$$ for all $$i, j$$, the cocycle condition $$f(ab) = \rho_V(a) f(b) + f(b) \rho_W(a)$$ doesn't constrain the values of $$f(x_i)$$ at all, so each choice of $$(c_1, \ldots, c_n) \in \mathbb{C}^n$$ yields a cocycle $$f_c$$ with $$f_c(x_i) = c_i \Id$$. Again $$B^1(V, V) = 0$$, so if $$c = [c_1 : \ldots : c_n]$$ is understood projectively, then the result of Problem 3.9.1(d) implies that for $$n > 1$$ we have an infinite family of nonisomorphic indecomposable extensions of the form $\rho(x_i) = \begin{pmatrix} 0 & c_i \\ 0 & 0 \end{pmatrix}$ where $$c$$ ranges over the infinite set $$\mathbb{P}\mathbb{C}^n$$.

Problem 3.9.3 Obviously all one-dimensional representations of $$P_Q$$ are irreducible. We claim that the converse is also true. Let $$V$$ be a nonzero representation of $$P_Q$$, so a vector space $$V_i$$ is associated to each vertex of $$Q$$, and $$V = \bigoplus_{i\in Q} V_i$$. Let $$Q'$$ be the set of vertices to which a nonzero vector space has been assigned. Let $$v$$ be some member of $$Q'$$ from which no other member of $$Q'$$ is reachable. Then $$V_v$$ is a subrepresentation of $$V$$ in which $$p_v$$ acts as the identity, $$p_w$$ acts as zero for all vertices $$w \in Q \setminus \{v\}$$, and $$a_e$$ acts as zero for all edges $$e \in Q$$. Each one-dimensional subspace of $$V_v$$ is itself a subrepresentation. Therefore, if $$V$$ is irreducible, then it must be one-dimensional.

The one-dimensional representations of $$P_Q$$ are obtained by choosing some vertex $$v \in Q$$ and assigning a one-dimensional vector space to $$v$$ and the zero vector space to all other vertices. Explicitly \begin{align*} \rho(p_w) &= \delta_{vw} \Id \\ \rho(a_e) &= 0 \end{align*} We will denote this representation by $$(V_v, \rho_v)$$. It is easy to see that $$V_v$$ and $$V_w$$ are nonisomorphic when $$v \ne w$$, as the former has $$\rho_v(w) = 0$$ and the latter has $$\rho_w(w) = \Id$$.

Let us now compute $$Z^1(V_w, V_v)$$. Let $$f$$ be a cocycle. Let $$x, y$$ be vertices and $$a$$ an edge. By the cocycle condition and the defining relations of $$P_Q$$, \begin{gather} f(\delta_{xy} p_x) = f(p_x p_y) = \rho_v(p_x) f(p_y) + f(p_x) \rho_w(p_y) = \delta_{vx} f(p_y) + \delta_{wy} f(p_x) \label{eqn:q1} \\ f(\delta_{a''v} a) = f(p_v a) = \rho_v(p_v) f(a) + f(p_v) \rho_w(p_v) = f(a) + \delta_{wv} f(p_v) \label{eqn:q2} \\ f(\delta_{a'w} a) = f(a p_w) = \rho_v(a) f(p_w) + f(a) \rho_w(p_w) = f(a) + \delta_{vw} f(p_w) \label{eqn:q3} \end{gather} We proceed by cases. First, suppose $$v \ne w$$. Then $$\pref{eqn:q2}$$ and $$\pref{eqn:q3}$$ become $f(a) = f(\delta_{a'' v}a) = f(\delta_{a'w}a)$ which implies that $$f(a) = 0$$ unless $$a$$ is an edge from $$w$$ to $$v$$. In $$\pref{eqn:q1}$$, substitute $$x = v, y = w$$ to obtain $0 = f(p_w) + f(p_v)$ Or, if $$x$$ is different from both $$v$$ and $$w$$, substitute $$y = w$$ to obtain $0 = f(p_x)$ Thus $$Z^1(\rho_v, \rho_w)$$ is parametrized by $$\lambda \in \Hom_k(V_w, V_v)$$ and $$\mu_a \in \Hom_k(V_w, V_v)$$ where $$a$$ ranges over the edges from $$w$$ to $$v$$, with $$f(\rho_w) = \lambda, f(\rho_v) = -\lambda$$, $$f(a) = \mu_a$$ if $$a$$ is an edge from $$w$$ to $$v$$, and $$f$$ vanishes on any other path.

If $$X \in \Hom_k(V_v, V_w)$$ then the coboundary $$\d X$$ is $$\d X(a) = \rho_W(a) X - X \rho_V(a)$$, implying \begin{align*} \d X(p_w) &= X \\ \d X(p_v) &= -X \\ \end{align*} and $$\d X(a) = 0$$ where $$a$$ is any other path. So two cocycles differing only in the choice of $$\lambda$$ become equivalent in $$\mathrm{Ext}^1(V_w, V_v)$$, so the latter is parametrized by the $$\mu_a$$'s alone, and has dimension $$d(w, v)$$ (the number of directed edges from $$w$$ to $$v$$).

Now consider the case where $$v = w$$. In $$\pref{eqn:q1}$$, substitute $$x = y = v$$ to obtain $f(p_v) = f(p_v p_v) = f(p_v) + f(p_v)$ therefore $$f(p_v) = 0$$. Or, if $$x$$ is different from $$v$$, substitute $$y = v$$ to obtain $0 = f(p_x p_v) = f(p_x)$ so $$f$$ vanishes for all vertices. Therefore, $$\pref{eqn:q2}$$ and $$\pref{eqn:q3}$$ become $f(a) = f(\delta_{a''v}a) = f(\delta_{a'v}a)$ Since $$Q$$ doesn't contain any oriented cycles, there is no self-edge from $$v$$, so at least one of $$\delta_{a''v}$$ and $$\delta_{a'v}$$ must be zero for all $$a$$. So in fact $$f$$ vanishes identically. Therefore, $$Z^1(V_v, V_v)$$ and $$\mathrm{Ext}^1(V_v, V_v)$$ are trivial.

The result that $$\mathrm{Ext}^1(V_v, V_v)$$ is trivial can actually be seen as a special case of the result that $$\mathrm{Ext}^1(V_w, V_v)$$ has dimension equal to $$d(w, v)$$, since $$d(v, v) = 0$$ in $$Q$$, although the proofs of the two cases were slightly different

We can now classify the two-dimensional representations of $$P_Q$$. Let $$V$$ be such a representation. It always has some one-dimensional subrepresentation, which is isomorphic to $$V_v$$ for some vertex $$v$$. The quotient $$V/V_v$$ is a one-dimensional representation, which is isomorphic to $$V_w$$ for some vertex $$w$$. $$\mathrm{Ext}^1(V_w, V_v)$$ has dimension equal to $$d(w, v)$$, the number of directed edges from $$w$$ to $$v$$. Let $$\varphi : V_w \to V_v$$ be an arbitrary nonzero linear map. The result of Problem 3.9.1(d) implies that all extensions of $$V_w$$ by $$V_v$$ are of the form $$U_{vwc}$$, parametrized by $$c = [c_1 : \ldots : c_{d(w, v)}] \in \mathbb{P}k^{d(w, v)}$$, with $\rho_{vwc}(a) = \begin{pmatrix} \rho_v(a) & f_{vwc}(a) \\ 0 & \rho_w(a) \end{pmatrix}$ where $$f_{vwc}$$ is the cocycle that assigns $$c_i \varphi$$ to the $$i$$th edge from $$w$$ to $$v$$.

Observe that $$U_{vwc}$$ has the property that $$\rho(p_x)$$ vanishes precisely when $$x$$ is different from both $$v$$ and $$w$$. This implies that if $$U_{vwc} \cong U_{v'w'c'}$$ then the two sets $$\{v, w\}$$ and $$\{v', w'\}$$ coincide. In the case where $$v = v'$$ and $$w = w'$$, the result of Problem 3.9.1(d) implies that $$U_{vwc} \cong U_{vwc'}$$ if and only if $$c = c'$$. We must still determine when $$U_{vwc}$$ and $$U_{wvc'}$$ can be isomorphic. If $$c = c' = 0$$, then $$U_{vwc} \cong V_v \oplus V_w$$ and $$U_{wvc'} \cong V_w \oplus V_v$$ so they are isomorphic. If $$c \ne 0$$, then there is at least one edge $$a$$ from $$w$$ to $$v$$ such that $$c_a \ne 0$$, implying $$\rho_{vwc}(a) \ne 0$$, but then there cannot be an edge from $$v$$ to $$w$$, so in $$U_{wvc'}$$ we must have $$\rho_{wvc}(a) = 0$$. So in this case $$U_{vwc} \not\cong U_{wvc'}$$. And similar reasoning shows that $$U_{vwc} \not\cong U_{wvc'}$$ whenever $$c' \ne 0$$. So the nonisomorphic two-dimensional representations of $$P_Q$$ are as follows:

• The direct sums $$V_v \oplus V_w$$ where $$v$$ and $$w$$ may be the same or different from each other;
• The indecomposable representations $$U_{vwc}$$ parametrized by $$c \in \mathbb{P}k^{d(w, v)} \setminus \{0\}$$, which exist only when there is at least one edge from $$w$$ to $$v$$.

The representations $$V_{vv}$$ assign a two-dimensional vector space to the vertex $$v$$ and the zero space to all vertices. The representations $$V_{vw}$$ with $$v \ne w$$ assign a one-dimensional vector space to $$v$$, a one-dimensional vector space to $$w$$, the zero space to all other vertices, and the zero map to all edges. The representations $$U_{vwc}$$ with $$c \ne 0$$ assign a one-dimensional vector space to each of $$v$$ and $$w$$, the zero space to all other edges, an arbitrary map to each edge from $$w$$ to $$v$$ as determined by the corresponding element of $$c$$ (but subject to the restriction that at least one edge has a nonzero map) and zero to all other edges; such representations are indecomposable and the $$V_v$$ subspace is clearly identifiable by reducing the space assigned to $$w$$ to the zero space. Notice that scaling all the edges from $$w$$ to $$v$$ results in an isomorphic representation since it has the same effect as simply rescaling the basis vector for the space assigned to either $$v$$ or $$w$$.

Problem 3.9.4

1. We will state some useful facts here without proof:

• If $$R$$ is any (possibly non-commutative) ring and $$R[[t]]$$ is the ring of formal power series with coefficients in $$R$$, and $$f, g \in R[[t]]$$, and $$n \in \mathbb{N}$$, then the coefficient of $$t^n$$ in $$fg$$ is determined by the coefficients of $$1, t, \ldots, t^n$$ in $$f$$ and $$g$$.
• If $$f \in R[[t]]$$ and the constant term of $$f$$ is a unit in $$R$$, then $$f$$ is a unit in $$R[[t]]$$, and for each $$n \in \mathbb{N}$$, the coefficient of $$t^n$$ in $$f^{-1}$$ is determined by the coefficients of $$1, t, \ldots, t^n$$ in $$f$$.

Now, suppose $$A$$ and $$(V, p)$$ are given as in the problem, with $$\mathrm{Ext}^1(V, V) = 0$$.

Suppose $$n \in \mathbb{N}$$ and $$n \ge 1$$. Suppose $$\tilde\rho = \rho + t\rho_1 + t^2\rho_2 + \ldots$$ is a deformation of $$\rho$$ such that the coefficient of $$t^i$$ vanishes identically (that is, for all $$a$$\) whenever $$1 \le i \le n-1$$; that is, $$\tilde\rho = \rho + t^n \rho_n + O(t^{n+1})$$. Then there exists $$b \in \End V$$ such that the coefficient of $$t^i$$ in $$(1 + t^n b)\tilde\rho (1 + t^n b)^{-1}$$ vanishes whenever $$1 \le i \le n$$.

We have $$\tilde\rho(ab) = \tilde\rho(a)\tilde\rho(b)$$ for all $$a, b \in A$$. That is, $$\rho(ab) + t^n \rho_n(ab) + O(t^{n+1}) = (\rho(a) + t^n \rho_n(a) + O(t^{n+1}))(\rho(b) + t^n \rho_n(b) + O(t^{n+1}))$$, and equating coefficients of $$t^n$$, this implies $\rho_n(ab) = \rho(a)\rho_n(b) + \rho_n(a)\rho(b)$ so $$\rho_n \in Z^1(V, V)$$. Since $$\mathrm{Ext}^1(V, V) = 0$$, this implies $$\rho_n \in B^1(V, V)$$, so there exists $$b \in \End V$$ such that $\rho_n = \rho b - b \rho$ Then \begin{align*} (1 + t^n b)\tilde\rho(1 + t^n b)^{-1} &= (1 + t^n b + O(t^{n+1}))(\rho + t^n \rho_n + O(t^{n+1})) (1 - t^n b + O(t^{n+1})) \\ &= (\rho + t^n(b\rho + \rho_n) + O(t^{n+1}))(1 - t^n b + O(t^{n+1})) \\ &= \rho + t^n(-\rho b + b\rho + \rho_n) + O(t^{n+1}) \\ &= \rho + O(t^{n+1}) \end{align*} as required.

Let $$\tilde\rho$$ be any given deformation of $$\rho$$. The Lemma implies that we can choose $$b_1 \in \End V$$ so that $$(1 + tb_1)\tilde\rho (1 + tb_1)^{-1}$$ has vanishing linear term. Then, the Lemma implies that we can choose $$b_2 \in \End V$$ so that $$(1 + t^2 b_2)(1 + tb_1)\tilde\rho (1 + tb_1)^{-1}(1 + t^2 b_2)^{-1}$$ has vanishing linear and quadratic terms, and so on. We will assume the axiom of dependent choice and assert that there is therefore an infinite sequence $$b_1, b_2, \ldots$$ with $$b_i \in \End V$$ such that for each $$n \in \mathbb{N}$$, the coefficients of $$t^1, \ldots, t^n$$ in $$(\prod_{i=1}^n 1 + t^i b_i)\tilde\rho (\prod_{i=1}^n 1 + t^i b_i)^{-1}$$ vanish identically.

Let $$b = \prod_{i=1}^\infty 1 + t^i b_i$$. This is well-defined because only factors with $$i \le n$$ contribute to the coefficient of $$t^n$$ in the product. Consider the representation $$\tilde{\tilde\rho} = b\tilde\rho b^{-1}$$. Clearly the constant term of $$\tilde{\tilde\rho}$$ is $$\rho$$. For $$n \ge 1$$, the coefficient of $$t^n$$ in $$\tilde{\tilde\rho}$$ is determined by $$\tilde\rho$$ and the coefficients of $$b$$ and $$b^{-1}$$ up to degree $$n$$. But $$b = O(t^{n+1}) + \prod_{i=1}^n 1 + t^i b_i$$ and $$b^{-1} = O(t^{n+1}) + (\prod_{i=1}^n 1 + t^i b^i)^{-1}$$, therefore the coefficient of $$t^n$$ in $$b\tilde\rho b^{-1}$$ is 0 by construction of the $$b_i$$'s. This holds for all $$n$$, therefore $$\tilde{\tilde\rho} = \rho$$ and we conclude that $$\rho$$ and $$\tilde\rho$$ are isomorphic.

2. Let $$A = \mathbb{C}[x]/x^2$$. This algebra is obtained by setting $$n = 1$$ in the class of algebras described in Problem 3.9.2(b). We found that such an algebra has a unique irreducible representation $$(V, \rho)$$, of dimension one, in which $$x$$ acts as the zero operator, and that the space $$\mathrm{Ext}^1(V, V)$$ has dimension $$n$$. In particular, for $$n = 1$$, $$\mathrm{Ext}^1(V, V)$$ is nonzero.

Let $$\tilde\rho = \rho + t \rho_1 + t^2 \rho_2 + \ldots$$ be a deformation of $$\rho$$. We will show that $$\rho_i = 0$$ for all $$i \ge 1$$. The proof is by induction. For the inductive case, suppose that $$\rho_1 = \rho_2 = \ldots = \rho_{i-1} = 0$$. Then we have $$\tilde\rho = \rho + t^i \rho_i + O(t^{i+1})$$. Now \begin{align*} 1 + t^i \rho_i(1) + O(t^{i+1}) &= \tilde\rho(1) = \tilde\rho(1 \cdot 1) = \tilde\rho(1)^2 \\ &= (1 + t^i \rho_i(1) + O(t^{i+1}))^2 \\ &= 1 + 2 t^i \rho_i(1) + O(t^{i+1}) \end{align*} which implies that $$\rho_i(1) = 0$$. Also \begin{align*} 0 &= \tilde\rho(0) = \tilde\rho(x^2) = \tilde\rho(x)^2 \\ &= (\rho(x) + t^i \rho_i(x) + O(t^{i+1}))^2 = (t^i \rho_i(x) + O(t^{i+1}))^2 \\ &= t^{2i} \rho_i(x)^2 + O(t^{2i+1}) \end{align*} therefore $$\rho_i(x)^2 = 0$$, and since $$\rho_i(x) \in \End V \cong \mathbb{C}$$, this means $$\rho_i(x)$$ is just 0. As $$\{1, x\}$$ is a basis of $$A$$, we have established that $$\rho_i = 0$$. For the base case, just put $$i = 1$$ and this argument still works because the assumptions that $$\rho_1 = \rho_2 = \ldots = \rho_{i-1} = 0$$ are vacuous.

So $$\mathrm{Ext}^1(V, V)$$ is nontrivial and yet all deformations of $$V$$ are trivial. The converse to (a) does not hold.

Problem 3.9.5

1. If $$\dim V = m$$, then a spanning set for $$\mathrm{Cl}(V)$$ is given by monomials of the form $$x_{i_1} x_{i_2} \ldots x_{i_p}$$ where $$1 \le i_1 < i_2 < \ldots < i_p \le m$$; the proof of this is not difficult, so we omit it, but the important consequence is that $$\dim\mathrm{Cl}(V) \le 2^m$$.

Suppose $$m = 2n$$ and the bilinear form $$\langle \cdot, \cdot \rangle$$ is nondegenerate. We use the fact that all nondegenerate symmetric bilinear forms can be diagonalized (we will not prove this here). Following the Hint, let $$e_1, \ldots, e_n, f_1, \ldots, f_n$$ be an orthnormal basis of $$V$$ with respect to $$\langle \cdot, \cdot \rangle$$, and define \begin{align*} a_i &= \frac{1}{2}(e_i + if_i) \\ b_i &= \frac{1}{2}(e_i - if_i) \end{align*} Then $$a_1, \ldots, a_n, b_1, \ldots, b_n$$ form a basis of $$V$$ in which $$\langle a_i, a_j \rangle = \langle b_i, b_j \rangle = 0$$ and $$\langle a_i, b_j \rangle = \delta_{ij}/2$$, as in the Hint. Let $$S$$ be the vector space $$\wedge(a_1, \ldots, a_n)$$. For each $$i$$, define \begin{align*} \rho(a_i)(\omega) &= a_i \wedge \omega \\ \rho(b_i)(\omega) &= \iota_{a_i} \omega \end{align*} and extend $$\rho$$ to an algebra homomorphism $$\rho : \mathbb{C}\langle a_1, \ldots, a_n, b_1, \ldots, b_n \rangle \to \End S$$. Observe that \begin{align*} \rho(a_i a_j + a_j a_i - 2\langle a_i, a_j \rangle)(\omega) &= \rho(a_i)\rho(a_j)(\omega) + \rho(a_j)\rho(a_i)(\omega) \\ &= a_i \wedge (a_j \wedge \omega) + a_j \wedge (a_i \wedge \omega) \\ &= (a_i \wedge a_j) \wedge \omega + (a_j \wedge a_i) \wedge \omega \\ &= (a_i \wedge a_j + a_j \wedge a_i) \wedge \omega \\ &= 0 \end{align*} and \begin{align*} \rho(b_i b_j + b_j b_i - 2\langle b_i, b_j \rangle)(\omega) &= \rho(b_i)\rho(b_j)(\omega) + \rho(b_j)\rho(b_i)(\omega) \\ &= \iota_{a_i} \iota_{a_j} \omega + \iota_{a_j} \iota_{a_i}(\omega) \\ &= 0 \end{align*} where we have used the identity $$\iota_X \iota_Y = -\iota_Y \iota_X$$, and \begin{align*} \rho(a_i b_j + b_j a_i - 2\langle a_i, b_j\rangle)(\omega) &= \rho(a_i)\rho(b_j)(\omega) + \rho(b_j)\rho(a_i)(\omega) - \delta_{ij}\omega \\ &= a_i \wedge \iota_{a_j}\omega + \iota_{a_j}(a_i \wedge \omega) - \delta_{ij}\omega \\ &= a_i \wedge \iota_{a_j}\omega + (\iota_{a_j}a_i) \wedge \omega - a_i \wedge \iota_{a_j}\omega - \delta_{ij}\omega \\ &= \delta_{ij} \omega - \delta_{ij}\omega \\ &= 0 \end{align*} where we have used the graded Leibniz identity $\iota_X(\omega_1 \wedge \omega_2) = (\iota_X \omega_1) \wedge \omega_2 + (-1)^k \omega_1 \wedge (\iota_X \omega_2)$ where $$\omega$$ is a $$k$$-form. Since $$\rho$$ vanishes on the ideal generated by the defining relations of $$\mathrm{Cl}(V)$$, we can restrict the domain of $$\rho$$ to $$\mathrm{Cl}(V)$$, forming a representation.

We now show that the representation $$(S, \rho)$$ is irreducible. Suppose $$\omega \in S \setminus \{0\}$$. Express $$\omega$$ in the basis of $$S$$ induced by $$a_1, \ldots, a_n, b_1, \ldots, b_n$$ and pick some $$k$$-form in $$S$$ of maximal grade with nonzero coefficient, say, $$c a_{i_1} \wedge \ldots a_{i_p}$$ where $$c \in \mathbb{C}$$. Then $$c^{-1} b_{i_p} \ldots b_{i_1} \omega = 1$$, so every nonzero subrepresentation of $$S$$ contains the element $$1$$. It is obvious that $$1$$ is cyclic in $$S$$, so $$S$$ is irreducible. The dimension of $$S$$ is $$2^n$$. By Theorem 3.5.4, $$\dim \mathrm{Cl}(V) - \dim \mathrm{Rad}(\mathrm{Cl}(V)) \ge 2^{2n}$$. Since $$\dim \mathrm{Cl}(V) \le 2^{2n}$$, this can only be satisfied if $$\dim \mathrm{Cl}(V) = 2^{2n}$$ and $$\mathrm{Rad}(\mathrm{Cl}(V)) = 0$$, so $$\mathrm{Cl}(V)$$ is semisimple and $$(S, \rho)$$ is its only unique irreducible representation.

In the case $$m = 2n + 1$$, let the orthonormal basis be $$e_1, \ldots, e_n, f_1, \ldots, f_n, c$$ and define $$a_i, b_i$$ as before. Then $$a_1, \ldots, a_n, b_1, \ldots, b_n, c$$ form a basis of $$V$$ with $$\langle a_i, a_j \rangle = \langle b_i, b_j \rangle = 0$$, $$\langle a_i, b_j\rangle = \delta_{ij}/2$$, $$\langle a_i, c\rangle = \langle b_i, c\rangle = 0$$, and $$\langle c, c \rangle = 1$$, as in the Hint. Let $$S = \wedge(a_1, \ldots, a_n)$$ as before. Define the maps $$\rho_+$$ and $$\rho_-$$ such that, when $$\omega$$ is a $$k$$-form, \begin{align*} \rho_\pm(a_i)(\omega) &= a_i \wedge \omega \\ \rho_\pm(b_i)(\omega) &= \iota_{a_i} \omega \\ \rho_+(c)(\omega) &= (-1)^k c \\ \rho_-(c)(\omega) &= (-1)^{k+1} c \end{align*} and extend $$\rho_\pm$$ to algebra homomorphisms $$\mathrm{C}\langle a_1, \ldots, a_n, b_1, \ldots, b_n, c\rangle \to \End S$$. $$\rho$$ vanishes for the defining relations involving only $$a$$'s and $$b$$'s as in the even case; also, where $$\omega$$ is a $$k$$-form, \begin{align*} \rho_+(c^2 - \langle c, c\rangle)(\omega) &= \rho_+(c)(\rho_+(c)(\omega)) - \omega \\ &= (-1)^k(-1)^k \omega - \omega \\ &= 0 \end{align*} and \begin{align*} \rho_+(a_i c + c a_i - \langle a_i, c \rangle)(\omega) &= \rho_+(a_i)(\rho_+(c)(\omega)) + \rho_+(c)(\rho_+(a_i)(\omega)) \\ &= a_i \wedge ((-1)^k \omega) + (-1)^{k+1}(a_i \wedge \omega) \\ &= (-1)^k a_i \wedge \omega + (-1)^{k+1}a_i \wedge \omega \\ &= 0 \end{align*} and \begin{align*} \rho_+(b_i c + c b_i - \langle b_i, c \rangle)(\omega) &= \rho_+(b_i)(\rho_+(c)(\omega)) + \rho_+(c)(\rho_+(b_i)(\omega)) \\ &= \iota_{a_i}((-1)^k \omega) + (-1)^{k-1}(\iota_{a_i}\omega) \\ &= (-1)^k\iota_{a_i}\omega + (-1)^{k-1}\iota_{a_i}\omega \\ &= 0 \end{align*} and similar relations hold for $$\rho_-$$, and both can be extended by linearity to all of $$S$$. As in the even case, $$\rho_+$$ and $$\rho_-$$ can therefore be restricted to $$\mathrm{Cl}(V)$$ and become representations of the latter, which we respectively denote $$S_+$$ and $$S_-$$.

The proof that $$S_\pm$$ are irreducible is the same as in the even case.

Define $$\gamma = c b_n \ldots b_2 b_1 a_1 a_2 \ldots a_n$$. Then $$\gamma$$ acts on $$S_+$$ as $$(-1)^n$$ times the projection onto the one-dimensional subspace generated by the unit element of $$S_+$$ while $$\gamma$$ acts on $$S_-$$ as $$(-1)^{n+1}$$ times the similar projection. Therefore $$\chi_{S_+}(\gamma) = (-1)^n$$ and $$\chi_{S_-}(\gamma) = (-1)^{n+1}$$. This establishes that $$S_+$$ and $$S_-$$ are nonisomorphic. Since $$S_+$$ and $$S_-$$ both have dimension $$2^n$$, a dimension-counting argument similar to that of the even case establishes that $$\mathrm{Cl}(V)$$ is semisimple and $$S_+$$ and $$S_-$$ are its only irreducible representations.

2. If $$\langle \cdot, \cdot \rangle$$ is nondegenerate, we proved in part (a) that the Clifford algebra is semisimple. If $$\langle \cdot, \cdot \rangle$$ is degenerate, form a basis $$x_1, \ldots, x_m$$ of $$V$$ such that $$\langle x_i, x_j \rangle = 0$$ when $$i \ne j$$, and $$\langle x_i, x_i \rangle = 1$$ for $$1 \le i \le d$$, and $$\langle x_i, x_i \rangle = 0$$ for $$d < i \le n$$, where $$d \in \{1, \ldots, n\}$$. Consider the ideal $$I_n = \langle x_n \rangle \subseteq A$$. Suppose $$a, b \in I_n$$. Then $$a, b$$ are linear combinations of terms of the form $$e x_n e'$$ and $$f x_n f'$$, where $$e, f, e', f' \in \mathrm{Cl}(V)$$. The product of these two terms contributes a term $$e x_n e' f x_n f'$$ to $$ab$$. Write $$e' f = \sum_{i=1}^p c_i g_i$$ where the $$g_i$$'s come from the spanning set of $$\mathrm{Cl}(V)$$ described in part (a), and $$c_i \in \mathbb{C}$$. Then $e x_n e' f x_n f' = \sum_{i=1}^p c_i e x_n g_i x_n f'$ Consider a single term in this sum, $$c_i e x_n g_i x_n f'$$. Note that $$x_n$$ anticommutes with $$x_1, \ldots, x_{n-1}$$, so if $$g_i$$ contains $$x_n$$ then we can swap the final $$x_n$$ with factors in $$g_i$$ to put the two $$x_n$$'s next to each other, showing that the product is 0. Otherwise, swapping the final $$x_n$$ with all the factors in $$g_i$$ puts the two $$x_n$$'s next to each other, so again the product is 0. Since this holds for all terms in $$e x_n e' f x_n f'$$, we conclude that the latter vanishes. Each term in the product $$ab$$ is of this form, so $$ab$$ vanishes. Since this holds for all $$a, b \in I_n$$, we conclude $$I_n^2 = 0$$, so $$\mathrm{Cl}(V)$$ is not semisimple.

By similar reasoning, if $$I = \langle x_{d+1}, \ldots, x_n\rangle$$ then we can show that $$I^{n-d+1} = 0$$. (The idea is to apply the pigeonhole principle; each term in a product of $$n-d+1$$ elements of $$I$$ must contain at least one of $$x_{d+1}, \ldots, x_n$$ at least twice, causing the term to vanish.) We pause here to prove a lemma that may seem obvious:

Let $$A$$ be a finite-dimensional algebra and let $$I \subseteq A$$ be a nilpotent (two-sided) ideal. If $$A/I$$ is semisimple, then $$I = \mathrm{Rad}(A)$$.

Let $$\pi : A \to A/I$$ be the quotient map. Suppose $$I'$$ is a nilpotent ideal of $$A$$. Then $$\pi(I')$$ is a nilpotent ideal of $$A/I$$. Since $$A/I$$ is semisimple, Proposition 3.5.3 implies that $$\pi(I')$$ is the zero ideal. Therefore $$I' \subseteq \ker\pi = I$$. So $$I$$ contains all nilpotent ideals of $$A$$. By Proposition 3.5.3, $$I = \mathrm{Rad}(A)$$.

Returning to the problem, the quotient $$\mathrm{Cl}(V)/I$$ is isomorphic to $$\mathrm{Cl}(V')$$ where $$V'$$ is the span of $$x_1, \ldots, x_d$$; the restriction of $$\langle \cdot, \cdot \rangle$$ is then the identity matrix on $$x_1, \ldots, x_d$$, so $$\mathrm{Cl}(V')$$ is semisimple. By the Lemma, $$\mathrm{Rad}(\mathrm{Cl}(V)) = I$$ and $$\mathrm{Cl}(V)/\mathrm{Rad}(\mathrm{Cl}(V)) \cong \mathrm{Cl}(V')$$.