Problem 3.9.1
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 \(f(1) = 0\).
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 \(\im \mathrm{d} \cong \Hom_k(W, V)/\Hom_A(W, V)\).
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 \(f' - f \in B^1(W, V)\).
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
Suppose \(f\) is a cocycle, \begin{equation} f(PQ) = \rho_b(P)f(Q) + f(P)\rho_a(Q) = P(b)f(Q) + Q(a)f(P) \label{eqn:coc} \end{equation} 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,
- The representations \(V_{ab} = V_a \oplus V_b\) where \(a, b \in \mathbb{C}^n\)
- 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.
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
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.
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
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.
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 \(\mathrm{Cl}(V)/\mathrm{Rad}(\mathrm{Cl}(V)) \cong \mathrm{Cl}(V')\).