# Exactness of the Chevalley-Eilenberg Resolution (Wednesday, April 21) :::{.remark} Recall that $\lieg$ was free over $k$ with an ordered basis \( \ts{ e_{\alpha} \st \alpha\in \Omega } \). We defined \[ V_n(\lieg) \da \Ug \tensor_k \Extalg^n \lieg \] with a differential \( d = \theta_1 + \theta_2 \). We claimed that $V_n(\lieg) \surjectsvia{\eps} k$ is a projective resolution, and we were showing that $V_*$ was an exact complex. ::: :::{.proof title="of theorem, continued"} We define a filtration \[ F_p V_n \da k \gens{ e_I \tensor e_{ \alpha_1} \wedge \cdots \wedge e_{\alpha_n} \st I = [\elts{\beta}{m}], \alpha_1 \leq \cdots \leq \alpha_n,\, m+n\leq p } .\] Note that $d: F_p V_n \to F_p V_{n-1}$, and in fact $\theta_2$ maps into $F_{p-1} V_{n-1}$. We'll focus on $\theta_1$ for simplicity. It lands in the same complex since we can rearrange elements in the sum defining the differential to express everything in terms of the given basis, where every expression will be of length one less. The commutation relation was \[ e_{ \beta} e_{\alpha} = e_{\alpha} e_{\beta} + [e_{\beta} e_{ \alpha} ] ,\] where the left-hand side is degree 2, and the right-hand side is a degree 2 term plus a degree 1 term. Moreover $d$ preserves the filtration: when rearranging, the degree $u$ term in $\theta_1$ will decrease to $m-1$, the expression following it may increase to $n+1$, and $(m-1) + (n+1) = m+n$. So $F_p V_*$ is a subcomplex of $V_*$, and we have \[ 0 = F_{-1} V_* \subseteq F_0 V_* \subseteq \cdots \subseteq F_p V_* \subseteq V_* = \Union_{p\geq 0} F_p V_* ,\] which is not a finite filtration, but is bounded below and exhaustive. So by the canonical convergence theorem (Weibel 5.5.1), there is a convergent spectral sequence \[ E_{p, q}^0 \da {F_p V_{p+q} \over F_{p-1} V_{p+q} } \abuts H_{p+q}(V_*(\lieg)) .\] We have $E_{p, q}^0 = 0$ unless - $p\geq 0$, since the exterior algebra is only graded in positive degrees. - $p+q = n \geq 0$ - $q\leq 0$, which requires some explanation. We have $m+n\leq p$, and so if $p+q=n \leq p-m$ for $m\geq 0$, $q\leq -m \leq 0$. So this is a 4th quadrant spectral sequence that is supported above the line $y=-x$. Recall that $E_{p, q}^1 = H_q^v(E_{p, *}^0)$. \begin{tikzcd} \bullet && \bullet \\ \\ \bullet &&& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&& \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&&& \bullet & \bullet & \bullet & \bullet & \bullet \\ &&&&&& \bullet & \bullet & \bullet & \bullet \\ &&&&&&& \bullet & \bullet & \bullet \\ &&&&&&&& \bullet \\ && \bullet \arrow[from=3-1, to=3-11] \arrow[from=1-3, to=9-3] \arrow[from=1-1, to=8-9] \arrow[from=3-5, to=4-5] \arrow[from=3-6, to=4-6] \arrow[from=4-6, to=5-6] \arrow["{d^0}", from=3-7, to=4-7] \arrow[from=4-7, to=5-7] \arrow[from=5-7, to=6-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Note that \[ E_{\infty}^0 \da F_0 V_0 / F_{-1} V_0 = k\tensor_k k \cong k ,\] since we take expressions with length zero in each factor defining $F_p V_n$. Moreover this position is already stable provided the $E_{p, 0}^{1} = 0$ for all $p$, the first and third quadrants are all zeros, and thus all differentials will be trivial from $E^1$ onward. :::{.claim} For $p>0$, $E_{p, *}^0$ is exact, and thus the spectral sequence collapses at $E^1$. ::: Note that turning the page yields \[ E_{p, q}^1 = \begin{cases} k & (p, q) = (0, 0) \\ 0 & \text{else}. \end{cases} \] Thus $H_n(V_*(\lieg)) = k$ in $n=0$ and zero elsewhere, which proves the result. :::{.proof title="Sketch"} For $q\gg 0$, define $A_q \da k \gens{ e_I \st I = [\elts{\beta}{q}] \text{ increasing} } \subseteq \Ug$. So $A_q$ is the $q$th graded piece of the standard increasing filtration by degree, \[ k = U_0 \subset U_1 \subset \cdots \subseteq \Ug .\] Note that this is a section standard filtration of $\Ug$ by degree with respect to the PBW basis[^exc_736]. We have $A_q \cong F_q V_0 / F_{q-1} V_0$ and \[ E_{p, q}^0 = {F_p V_{p+q} \over F_{p-1} V_{p+q} } \cong A_{-q} \tensor_k \Extalg^{p+q}\lieg .\] The negative sign is introduced since this is nonzero precisely when $-p\leq q \leq 0$ so $q$ is negative and $-q$ is positive. Now using the definition of $d: V_n \to V_{n-1}$, $d^0$ is vertical and \[ d^0: E_{p, q}^0 &\to E_{p, q-1}^0 \quad\quad n = p+q \\ \\ \cong d^0: A_{-1} \tensor_k \Extalg^n\lieg &\to A_{-q+1}\tensor_k \Extalg^{n-1} \lieg \quad\quad n = p+q .\] Recalling how $d^0$ was defined, note that we're modding out by lower order terms and thus brackets get killed when we commute elements to order them. \ By Weibel 7.3.6, $A \da \bigoplus_{q\geq 0} A_q$ is in fact a graded algebra, and $A \cong \gr \Ug$, the associated graded of $\Ug$. This turns out to be a polynomial ring on the indeterminates $\vector{x} = \ts{ e_{ \alpha } }_{\alpha\in \Omega}$, i.e. $A\cong k[\vector x]$. In Weibel section 4.5, Weibel studies the *Koszul* complex and the map $A \tensor_k \Extalg^* \lieg \to A$. By comparing the formula for $d$ between these two complexes, one observes that the Koszul complex differentials are equal to the $d^0$ here. So we have an equality of complexes \[ A \tensor_k \Extalg^* \lieg = \bigoplus _{p\geq 0} E_{p, *}^0 .\] Weibel section 4.5 shows that when $A\in \CRing$ with no zero divisors, e.g. a polynomial ring, then \[ H_n \qty{ A \tensor_k \Extalg^* \lieg } = \begin{cases} k & n=0 \\ 0 & \text{else}. \end{cases} \] On the other hand, we have \[ H_n \qty{ A \tensor_k \Extalg^* \lieg } &= \bigoplus H_{n-p}^v( E_{p, *}^0 ) \hspace{4em} p+q=n \implies q=n-p \\ &= \bigoplus _{p\geq 0} E_{p, n-p}^1 .\] But we've already shown that $E_{0, 0}^1 = k$, so all of the other $E^1$ terms must be zero. [^exc_736]: See exercise 7.3.6. ::: ::: :::{.remark} See section 4.5 on Koszul complex. We'll do 7.8 next time. :::