# Poincaré duality for (co)homology of Lie algebras?

Let $$R$$ be a commutative ring and $$\mathfrak{g}$$ a Lie $$R$$-algebra that has an $$R$$-module basis with $$n$$ elements.

In Algebra, Geometry, and Software Systems by Joswig & Takayama on p.200, it says that $$H_k(\mathfrak{g};R)\cong H_{n-k}(\mathfrak{g};R)$$ when $$R$$ is a field of characteristic $$0$$.

Does Poincaré duality $$H^k(\mathfrak{g};R)\cong H_{n-k}(\mathfrak{g};R)$$ hold over any $$R$$ (or at least any PID such as $$\mathbb{Z}$$)?

Also, the literature for (co)homology of Lie algebras seems scarce, please list any book that deals with this topic. So far, I have Weibel (Homological Algebra), Azcarraga & Izquierdo (Lie Groups, Lie Algebras, Cohomology), Hilgert & Neeb (Structure and Geometry of Lie Groups).

• Also on MathSE.
– Leo
Oct 16 '14 at 13:20
• For the literature, Fuks, Cohomology of Infinite-Dimensional Lie Algebras, despite the title, serves as a good general reference (for cohomology with coefficients in a field of zero characteristic, anyway). Oct 16 '14 at 15:50
• @YCor: It's Poincaré duality, so you should be looking at $H^2(g,R)$, which is indeed $(R/2R)^2$. Oct 17 '14 at 12:59
• It seems to me that the first place where it's done over arbitrary fields (for $\mathfrak{g}$ unimodular) is Koszul's paper (quoted in Dietrich Burde's answer), §III.7.
– YCor
Apr 14 '20 at 18:28

First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago).

Hazewinkel's most general result (for not necessarily trivial coefficients) is over a field. A result that he proves over a ring is as follows:

Let $$R$$ be a hereditary ring, $$\mathfrak{g}$$ an Lie algebra over $$R$$ which is a free $$R$$-module of rank $$n$$, and $$M$$ a finitely generated $$\mathfrak{g}$$-module which is projective as an $$R$$-module. Then there exists a canonical splitting exact sequence $$0\to\mathrm{Ext}^1(H^{s+1}(\mathfrak{g},M),R)\to H^{n-s}(\mathfrak{g},(M^{\mathrm{tw}})^*)\to H^s(\mathfrak{g},M)^*\to 0$$ where the $$\mathfrak{g}$$-module $$M^{\mathrm{tw}}$$ has the same underlying $$R$$-module, but the action on it is given by $$\rho^{\mathrm{tw}}(x)=\rho(x)-\mathop{\mathrm{tr}}(\mathop{\mathrm{ad}}(x))\cdot1$$.

In the case of $$M=R$$, $$(M^{\mathrm{tw}})^*$$ is $$R$$ with the $$\mathfrak{g}$$-action given by multiplying by trace of the adjoint action. The example of YCor in the comment to the original question, in particular, is addressed by this, of course (there, incidentally, $$R^{\mathrm{tw}}\simeq R$$ as $$\mathfrak{g}$$-modules): we have $$H^1(\mathfrak{sl}_2,\mathbb{Z})=0$$ and $$H^2(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$$, also we have $$H_1(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$$, and applying Hazewinkel's result for $$s=1$$ (so $$s+1=2$$), we see no contradiction with $$\mathrm{Ext}^1(H^{s+1}(\mathfrak{sl}_2,\mathbb{Z}),\mathbb{Z})=\mathrm{Ext}^1((\mathbb{Z}/2\mathbb{Z})^2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$$.

However, for the kind of information you seek (Poincaré-type duality between homology and cohomology), the answer is simpler, since the main reason for the universal coefficients-kind phenomena in the formulas of Hazewinkel is that we need to dualise something we do not want to dualise :)

Let me actually first give an example similar to the one Mariano gave, it is a nice simple exercise anyway, and it shows once again that twisting coefficients is unavoidable. Let $$\mathfrak{g}$$ be a Lie algebra over $$\mathbb{Z}$$ with two basis elements $$x,y$$ and the bracket $$[x,y]=2y$$. Then $$H_i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ 0, i>1,\end{cases}$$ $$H_i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}\mathbb{Z}/2\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases}$$ $$H^i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}/2\mathbb{Z}, i=2,\\ 0, i>2,\end{cases}$$ $$H^i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}0, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases}$$

This example makes one think that we must have $$H_i(\mathfrak{g},R)\simeq H^{n-i}(\mathfrak{g},R^{\mathrm{tw}})\quad \text{ and }\quad H_i(\mathfrak{g},R^{\mathrm{tw}})\simeq H^{n-i}(\mathfrak{g},R).$$ Indeed, this is true because the pairs of complexes computing that (co)homology are isomorphic. Basically, take some basis $$e_1\ldots,e_n$$ of $$\mathfrak{g}$$, and send each multivector $$e_{i_1}\wedge\cdots\wedge e_{i_k} (i_1<\dots to the skew-symmetric multilinear function $$\phi$$ for which, whenever $$j_1<\dots, we have $$\phi(e_{j_1},\ldots,e_{j_l})=\delta_{I,\{1,\ldots,n\}\setminus J}$$. This gives an isomorphism of the Chevalley-Eilenberg complexes computing the respective (co)homology groups. (The computation that this takes one differential into another (up to a sign) needs some careful bookkeeping but is quite straightforward, and is essentially contained between the lines in Hazewinkel's paper).

• Thank you, this is the kind of answer I was looking for!
– Leo
Oct 20 '14 at 20:10

As far as I know, a generalisation of Poincare duality for Lie algebra cohomology over rings is given in

M. Hazewinkel, "A duality theorem for the cohomology of Lie algebras" Math. USSR-Sb. , 12 (1970) pp. 638–644.

There is a large literature on cohomology of Lie algebras. Here is a very short list of articles and books:

C. Chevalley, S. Eilenberg, Cohomology theory of Lie groups and Lie algebras, Trans. Amer. Math. Soc. 63 (1948), 85-124.

G. P. Hochschild, J.-P. Serre, Cohomology of Lie algebras, Ann. of Math. 57 (1953), 591-603.

J. L. Koszul, Homologie et cohomologie des algèbres de Lie, Bull. Soc. Math. France , 78 (1950) pp. 65–127

J. C. Jantzen, Representations of Algebraic groups, Pure and Applied Mathematics, vol. 131, Boston, etc., 1987 (Academic).

J. C. Jantzen, Restricted Lie algebra cohomology, Lecture Notes in Math. 1271 (1986), 91-108.

A. W. Knapp, Lie groups, Lie algebras and cohomology, Mathematical Notes, Princeton University Press, 1988, 509 pp.

• If I remember correctly, Hazewinkel's paper is dealing with a different question. It is more general in that it proves some sort of duality for any module of coefficients, not just the ground ring, but the result is proved only in the case of algebras which are free as $R$-modules. I doubt that this is what the OP wanted. Oct 16 '14 at 21:15
• @VladimirDotsenko, the OP does consider only Lie algebras free over the base ring. Oct 16 '14 at 22:12
• @MarianoSuárez-Alvarez: you are right! I did not read it carefully. Oct 17 '14 at 6:49

Let $g$ be the $3$-dimensional complex Lie algebra with basis $\{x,y,z\}$ and $[x,y]=y$, $[x,z]=y+z$ and $[y,z]=0$. Then $H_0(g,\mathbb C)=\mathbb C$ and $H^3(g,\mathbb C)=0$, if I computed correctly. This Lie algebra is not unimodular; if you add that hypothesis, then it should be true over fields at least.

Indeed, in that case the enveloping algebra is $n$-Calabi-Yau, and one has $HH^k(\mathcal U(g),M)\cong HH_{n-k}(\mathcal U(g),M)$ for all $\mathcal U(g)$-bimodules $M$, with $HH$ denoting Hochschild homology and cohomology. If you take $M=\hom_{\mathbb C}(\mathbb C,N)$ with $N$ a left $g$-module, with its usual bimodule structure, then this isomorphism becomes $H^k(g,N)\cong H_{n-k}(g,N)$.

If the Lie algebra is not unimodular, then one only has $$H^k(\mathcal U(g),M)\cong HH_{n-k}(\mathcal U(g),M\otimes_{\mathcal U(g)}\mathcal U(g)_\chi)$$ with $\mathcal U(g)_\chi$ a certain «twisting bimodule» constructed from the modular character of $g$, and this gives only a twisted duality for Lie algebra (co)homology.

• For a Lie algebra free over a commutative base ring, I'd guess exactly the same thing holds provided one is talking about Lie algebra cohomology relative to the base ring. Oct 16 '14 at 22:34
• A Lie algebra $g$ is unimodular if $\operatorname{ad}_X$ has zero trace for all $X\in g$. Oct 16 '14 at 22:37
• Are there any equivalent characterizations of unimodular? For example, let $\preceq$ be a partial ordering of $\{1,\ldots,n\}$. Are $\mathfrak{gl}^\preceq=\{a\in\mathfrak{gl}_n; a_{ij}=0\text{ for }i\npreceq j\}$ and $\mathfrak{gl}^\prec=\{a\in\mathfrak{gl}_n; a_{ij}=0\text{ for }i\nprec j\}$ unimodular? Special cases (when $\prec$ is a linear ordering) are the Lie algebras of all (strictly) upper triangular matrices. Does PD hold for any unimodular LA over any ring?
– Leo
Oct 16 '14 at 22:48
• Have tou tried computing the traces in your two examples? There is no difficulty in doing that, really. Oct 16 '14 at 23:39
• Incidentally, the paper of Hazewinkel quoted in the other answer tells exactly that: to obtain duality, one twists a representation by replacing $\rho(x)$ by $\rho(x)-\mathop{\mathrm{tr}}(\mathop{\mathrm{ad}}(x))1$. Oct 17 '14 at 6:48