## 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:

- 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, \(s \in I\).
- 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 \(I\).
- 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 \(I\).
- 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- 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 \(h_A(t) = h_B(t)^m = (1-t)^{-m}\).
- There are \(m^n\) words of length \(n\), so \(h_A(t) = 1 + mt + m^2 t^2 + \ldots = \frac{1}{1-mt}\).
- 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.
- 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 \(h_A(t) = \sum_n \tr(O M_Q^n) t^n = \tr\left(O \sum_n t^n M_Q^n\right) = \tr\left(O(I - tM_Q)^{-1}\right)\).