Brian Bi

## Section 2.8. Quivers

Problem 2.8.6 For clarity, use the name $$A$$ to refer to the free algebra over the generators $$p_i, a_h$$ $$(i \in I, h \in E)$$, as described in the problem, and let $$I \subseteq A$$ denote the ideal generated by the relations defined in the problem, i.e., $$p_i^2 - p_i, p_i p_j, a_h p_{h'} - a_h, a_h p_j, p_{h''} a_h - a_h, p_i a_h$$. It is obvious that we are meant to identify $$p_i \in A$$ with the path in $$P_Q$$ consisting of only the vertex $$i$$, and $$a_h \in A$$ with the path in $$P_Q$$ consisting of only the edge $$h$$. We need to show that this induces a unique algebra isomorphism $$\varphi': A/I \to P_Q$$.

Let $$\varphi : A \to P_Q$$ be the homomorphism with $$\varphi(p_i) = i, \varphi(a_h) = h$$ for all $$i \in I, h \in E$$. It's easy to see that if $$(h_1, h_2, \ldots, h_n)$$ form a path in $$Q$$, then $$\varphi(a_{h_n} a_{h_{n-1}} \ldots a_{h_1})$$ gives the element in $$P_Q$$ corresponding to this path. Therefore $$\varphi$$ is surjective.

We claim that $$\ker\varphi = I$$. We prove this in two steps: first we show that $$I \subseteq \ker\varphi$$ and then that $$\ker\varphi \subseteq I$$. For the first step, it suffices to show that each defining relation $$r$$ is contained in $$\ker\varphi$$ because if $$\varphi(r) = 0$$ then $$\varphi(arb) = 0$$ automatically and so on by linearity. This is easy to verify for each of the defining relations: for example, if $$r = p_i^2 - p_i$$ then $$\varphi(r) = \varphi(p_i)^2 - \varphi(p_i)$$, but $$\varphi(p_i) = \varphi(p_i)^2 = i$$ so $$\varphi(r) = 0$$.

For the reverse direction, consider a word in $$A$$, $$w = g_1 g_2 \ldots g_n$$, where the $$g$$'s are generators of $$A$$. We know that $$\varphi(w)$$ consists of the concatenation of $$\varphi(g_n), \varphi(g_{n-1}), \ldots, \varphi(g_1)$$ when such concatenation is possible, and zero otherwise; in the former case $$\varphi(w)$$ will be nonzero and equal to a single path in $$P_Q$$. If $$\varphi(w) = 0$$ then $$w$$ contains a sequence of unconcatenatable factors $$s$$, that is, $$w = asb$$, under one of the following cases:

1. Two vertices that can't be concatenated because they are unequal. This takes the form $$s = p_i p_j$$ where $$i \neq j$$. Clearly, $$s \in I$$.
2. An edge can't be followed by a vertex because the vertex isn't the edge's target. This takes the form $$s = p_i p_{h''}^m a_h$$ where $$p_i \neq h''$$ and $$m \geq 0$$. If $$m \geq 1$$ then $$s$$ contains the factor $$p_i p_{h''}$$ and is therefore in $$I$$. If $$m = 0$$ then $$s$$ contains the factor $$p_i a_h$$ and is therefore in $$I$$.
3. A vertex can't be followed by an edge because the vertex isn't the edge's source. This takes the form $$s = a_h p_{h'}^m p_j$$ where $$j \neq h'$$. If $$m \geq 1$$ then $$s$$ contains the factor $$p_i p_j$$ and is therefore in $$I$$. If $$m = 0$$ then $$s$$ contains the factor $$a_h p_j$$ and is therefore in $$I$$.
4. Two edges that can't be concatenated because the first one's target is not the second one's source. This takes the form $$s = a_h p_{h'}^m p_{k''}^n a_k$$ where $$m, n \geq 0$$, $$h, k \in E$$ and $$h' \neq k''$$. If $$m, n > 0$$, then $$s \in \langle p_{h'} p_{k''} \rangle \subseteq I$$. Otherwise, $$s = a_h p_{h'}^{m+1} p_{k''}^{n+1} a_k + a_h (p_{h'}^m - p_{h'}^{m+1}) p_{k''}^{n+1} a_k + a_h p_{h'}^m (p_{k''}^n - p_{k''}^{n+1}) a_k$$ and each term lies in $$I$$.

Now let $$a \in A$$ be a linear combination of words and suppose that $$\varphi(a) = 0$$. We can write $$a = a' + z$$ where $$z$$ contains all the words in $$a$$ that map to 0 and $$a'$$ contains no such words. In $$a'$$, let us collect like terms, where two words are like if they map to the same path in $$Q$$. That is, $$a = z + \sum_P a_P$$ where each $$a_P$$ maps to a multiple of the path $$P$$ in $$Q$$ and all $$P$$ are distinct. Since the paths are linearly independent, we must have $$\varphi(a_P) = 0$$ for each $$P$$. Consider one $$a_P$$, which can be written as $$\sum_k c_k w_k$$ where each $$c$$ is a scalar and each $$w$$ a word. Obviously, it must be that $$\sum_k c_k = 0$$ in order for $$\varphi(a_P) = 0$$ to hold since each $$w_k$$ maps to $$P$$ and not a multiple of $$P$$. Let $$v$$ denote the canonical representation of the path $$P$$, which is defined as follows: if $$P$$ is a single vertex $$i$$, then $$v$$ is $$p_i$$, otherwise if $$P$$ is the path with edges $$(h_1, \ldots, h_n)$$ then $$v = a_{h_n} \ldots a_{h_1}$$. Then write $$a_P = \sum_k c_k v + \sum_k c_k (w_k - v) = \sum_k c_k (w_k - v)$$. We now prove that $$w_k - v \in I$$ for each $$w_k$$.

If $$P$$ is a single vertex $$i$$ then $$w_k = p_i^m$$ for some $$m \geq 0$$ and $$v = p_i$$. In this case it's easy to see that $$w_k - v \in I$$. Otherwise, write $$P$$ as the sequence of edges $$(h_1, \ldots, h_n)$$ and $$w_k$$ must take the form \begin{equation*} w_k = p_{h_n''}^{e_n} a_{h_n} p_{h_n'}^{e_{n-1}} a_{h_{n-1}} p_{h_{n-1}'}^{e_{n-2}} \ldots p_{h_1''}^{e_1} a_{h_1} p_{h_1'}^{e_0} \end{equation*} where each $$p$$ matches the surrounding edges in the obvious way, and each $$e$$ is greater than or equal to zero. We can successively eliminate all the $$p$$'s by adding elements that belong to $$I$$, for example, if $$e_n > 0$$, then we can add $$(p_{h_n''} - p_{h_n''}^{e_n}) a_{h_n} \ldots \in I$$ and then $$(a_{h_n} - p_{h_n''} a_{h_n}) p_{h_n'}^{e_{n-1}} a_{h_{n-1}} \ldots \in I$$ to $$w_k$$ to eliminate the factor of $$p_{h_n''}^{e_n}$$; repeating to eliminate all the $$p$$'s yields the desired result that $$w_k - v \in I$$.

Having obtained this result, we can now be assured that $$a \in \ker\varphi$$ implies $$a \in I$$, and therefore that $$I = \ker\varphi$$. By the first isomorphism theorem, $$\varphi$$ factors through $$A/I$$ to give the desired isomorphism $$\varphi' : A/I \to P_Q$$.

Problem 2.8.11
1. Since the degree of $$A[n]$$ is precisely the sum over all products of the form $$\deg B[i_1] \ldots \deg B[i_m]$$ with $$i_1, \ldots, i_m \geq 0$$ and $$i_1 + \ldots + i_m = n$$, where $$B$$ is the polynomial algebra over one variable, it follows that $$h_A(t) = h_B(t)^m = (1-t)^{-m}$$.
2. There are $$m^n$$ words of length $$n$$, so $$h_A(t) = 1 + mt + m^2 t^2 + \ldots = \frac{1}{1-mt}$$.
3. It is well known that $$A[n] = \binom{m}{n}$$ for the exterior algebra, which we will not prove here. So $$h_A(t) = \binom{m}{0} + \binom{m}{1}t + \binom{m}{2}t^2 + \ldots = (1+t)^m$$ by the binomial theorem.
4. If $$M_Q$$ is the adjacency matrix of $$Q$$, then the number of paths of length $$n$$ from vertex $$i$$ to vertex $$j$$ is given by $$(M_Q)^n_{ij}$$. Therefore $$A[n] = \tr(O M_Q^n)$$ where $$O$$ is the matrix of all ones with size equal to the number of vertices in $$Q$$. Then $$h_A(t) = \sum_n \tr(O M_Q^n) t^n = \tr\left(O \sum_n t^n M_Q^n\right) = \tr\left(O(I - tM_Q)^{-1}\right)$$.