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This question is about the relationship (rather, whether there is or ought to be a relationship) between torsion for the cohomology of certain Lie algebras over the integers, and torsion for associated Lie groups.

Let $\mathfrak{g}$ be a complex simple Lie algebra. Then we can choose a basis of Chevalley generators for $\mathfrak{g}$. The structure constants describing the Lie brackets among the Chevalley generators are all integers, so the Chevalley basis spans a Lie algebra over the integers, which I shall denote by $\mathfrak{g}_\mathbb{Z}$. We can compute the (Chevalley-Eilenberg) Lie algebra cohomology of $\mathfrak{g}_\mathbb{Z}$ with coefficients in $\mathbb{Z}$ using the usual Koszul complex, and thus get for each $1 \leq n \leq \dim \mathfrak{g}$ a finitely-generated abelian group $H^n(\mathfrak{g}_\mathbb{Z},\mathbb{Z})$.

On the other hand, (as I understand it) the Lie algebra cohomology of $\mathfrak{g}$ with coefficients in $\mathbb{C}$, denoted $H^*(\mathfrak{g},\mathbb{C})$, is the same as the cohomology of some associated (compact?) connected Lie group $G$. This philosophy is described in the section "Motivation" of the Wikipedia article on Lie algebra cohomology. (I'm not an expert on this matter, so I invite someone more knowledgable to correct me if I've got something incorrect. I think that the associated Lie group $G$ will have $\mathfrak{g}$ as its Lie algebra, or perhaps $\mathfrak{g}$ will be the complexification of the Lie algebra of $G$.) We can also compute $H^n(G,\mathbb{Z})$, the cohomology of $G$ with coefficients in $\mathbb{Z}$.

Is there, or ought there to be, a connection between the torsion of $H^n(G,\mathbb{Z})$ and the torsion of $H^n(\mathfrak{g}_\mathbb{Z},\mathbb{Z})$? Should the torsion primes of the two abelian groups be the same? Should the torsion primes of one be a subset of the torsion primes for the other?

To the best of my knowledge, the torsion primes of the compact connected simple Lie groups were worked out in the 1950s and/or 1960s, especially by Armand Borel, but at present no such list seems readily available for the Lie algebra cohomology side of the picture.

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A related question: mathoverflow.net/questions/11703/… –  algori Jun 4 '11 at 4:12

2 Answers 2

I don't know the answer to the actual question but here is a situation which should be similar but simpler. Consider an integral polynomial group law $G$, i.e., a group scheme structure on the affine space over $\mathbb Z$. We then have a torsion free finitely generated nilpotent group $G(\mathbb Z)$ and an integral Lie algebra $\mathfrak g_{\mathbb Z}$. We also have a compact manifold $X:=G(\mathbb R)/\mathbb(\mathbb Z)$ and as $G(\mathbb R)$ is contractible we have that the cohomology of $X$ is equal to the cohomology of $G(\mathbb Z)$ (as $X$ is a $K(G(\mathbb Z),1)$). We also have that $G(\mathbb Q)$ is the Malcev completion of $G(\mathbb Z)$ and (consequently) we have that the inclusion $G(\mathbb Z)\hookrightarrow G(\mathbb Q)$ induces an isomorphism on cohomology with rational coefficients. For more obvious reasons (the Chevalley-Eilenberg complexes are isomorphic) $\mathfrak g_{\mathbb Z}\hookrightarrow \mathfrak g_{\mathbb Q}$ induces an isomorphism in cohomology.

Furthermore, $\mathfrak g_{\mathbb Q}$ is the Lie algebra that correspond to the torsion free, divisible nilpotent group $G(\mathbb Q)$ under the Malcev correspondence (i.e., the Campbell-Baker-Hausdorff formula) which results in an isomorphism between the Lie algebra cohomology (with rational coefficients) of $\mathfrak g_{\mathbb Q}$ and the group cohomology (again with rational coefficients) of $G(\mathbb Q)$. Combining everything we get an isomorphism between the cohomology of $\mathfrak g_{\mathbb Z}$ and of $G(\mathbb Z)$ (equal to that of $X$) both with rational coefficients. If I remember correctly Larry Lambe computed examples where the torsion of the integral cohomology was different (perhaps even for strictly upper triangular $4\times4$-matrices).

We can approach the comparison somewhat differently: To begin with, there is a third cohomology involved, the cohomology of the group scheme (over $\mathbb Z$) $G$. We can put all three of these cohomologies on an equal footing. We may consider the subcomplex of the standard complex for computing group cohomology consisting of polynomial cochains $G(\mathbb Z)\hookrightarrow \mathbb Z$, i.e., as $G$ as a scheme is just affine space $G(\mathbb Z)=\mathbb Z^m$ for some $m$ and we demand that the cochain $\mathbb Z^{mn}\hookrightarrow \mathbb Z$ be given by a polynomial. This is clearly a subcomplex and computes the group scheme cohomology of $G$. We may consider another subcomplex consisting of the functions given by polynomials with rational coefficients (still inducing a map $G(\mathbb Z)\hookrightarrow \mathbb Z$). It again is a subcomplex which has been shown to compute the cohomology of the group $G(\mathbb Z)$ (i.e., the inclusion into all cochains is a quasi-isomorphism). Note that such functions are sums of products of binomal polynomials in the projection functions (of $\mathbb Z^{mn}$ to its factors). The third complex is not quite a subcomplex consisting it does of rational valued functions that can be written as sums of products of divided powers of the projection functions. I have never completely checked all the details but am pretty certain that this complex computes the Lie algebra cohomology of $\mathfrak g_{\mathbb Z}$.

As polynomials with integer coefficients are both polynomials with rational coefficients and divided power polynomials we get an explicit map from group scheme cohomology to both group cohomology of integral points and to the integral Lie algebra cohomology. Tensoring with $\mathbb Q$ induces isomorphisms of complexes giving an explicit realisation of the rational isomorphisms. Note however that the group scheme cohomology is definitely distinct from the Lie algebra cohomology and the group cohomology already in the case of the additive group scheme: The $2$-cocycle $((x+y)^p-x^p-y^p)/p$, $p$ a prime, is polynomial which is not a polynomial coboundary. It is however the coboundary of the cochain $(x^p-x)/p$ which is integer-valued as well as of the divided power cochain $x^p/p$.

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There is a simple observation about some odd behaviour of the prime $2$. The Lie group $SL_2(\mathbb{C})$ has torsion-free cohomology, because as a space it is weakly equivalent to $S^3$.

Consider the Lie algebra $\mathfrak{sl}_2(\mathbb{Z})$ of $2\times 2$-matrices with trace zero, and the usual commutator. This is a $\mathbb{Z}$-form of $\mathfrak{sl}_2$. Since the Cartan acts with weight $2$, the reduction of the above $\mathbb{Z}$-form to $\mathbb{F}_2$ is in fact not simple but solvable - the Cartan is an ideal and the quotient is abelian. This is still visible for the $\mathbb{Z}$-form, which has abelianization isomorphic to $(\mathbb{Z}/2)^{\oplus 2}$, generated by the matrices $$ \left(\begin{array}{cc} 0&1\\0&0\end{array}\right) \qquad\textrm{ and }\qquad \left(\begin{array}{cc} 0&0\\1&0\end{array}\right). $$ The universal coefficient formula then implies $H^2(\mathfrak{sl}_2(\mathbb{Z}),\mathbb{Z})\cong(\mathbb{Z}/2)^{\oplus 2}$.

In this case, $2$ would be a torsion prime for the Lie algebra, but not for the Lie group. For all other primes, the cohomology of the Lie algebra and of the Lie group coincide. At least the reason for this behaviour (the weight $2$ action of the Cartan on the weight spaces) persists for all the $\mathfrak{sl}_n$, but I have not made the calculations to see if this really implies the existence of non-trivial $2$-torsion in the homology of $\mathfrak{sl}_n$, $n\geq 3$.

Whether or not a relation should exist between torsion in the homology of the Lie algebra and the Lie group seems to be related to the mod $p$ splitting behaviour of the Lie group. In some cases, the mod $p$ homotopy type of the Lie group splits as a product of spheres. In these cases, I would expect stronger relationships (maybe excluding the odd behaviour of the prime $2$). However, I do not really know how to possibly relate these things - the $\mathbb{Z}$-form of the Lie algebra is related to differential forms, the splitting of the Lie groups is serious algebraic topology, and these only seem to interact nicely with rational coefficients via rational homotopy theory.

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