The length of this question has got a little bit out of hand. I apologize.

Basically, this is a question about the relationship between the cohomology of Lie groups and Lie algebras, and maybe periods.

Let $G$ be a complex reductive (connected) Lie group and let $T$ be a maximal torus of $G$. Set $\mathfrak{g}=Lie(G)$ and $\mathfrak{t}=Lie(T)$. Notice that $\mathfrak{t}$ has a natural integral structure: $\mathfrak{t}=\mathfrak{t}(\mathbf{Z})\otimes_\mathbf{Z}\mathbf{C}$ where $\mathfrak{t}(\mathbf{Z})$ is formed by all $x$ such that $\exp(2\pi ix)$ is the unit $e$ of $G$.

All there is to know about $G$ can be extracted from the (covariant) root diagram of $G$, which is formed by $\mathfrak{t}(\mathbf{Z})$, the sublattice $M$ corresponding to the connected component of the center of $G$ (this is a direct summand) and the coroot system $R$ of $G$, which is included in some complementary sublattice of $\mathfrak{t}(\mathbf{Z})$. For example, $\pi_1(G)$ is the quotient of $\mathfrak{t}(\mathbf{Z})$ by the subgroup spanned by $R$. See e.g. Bourbaki, Groupes et alg`ebres de Lie IX, 4.8-4.9 (Bourbaki gives a classification in terms of compact groups, but this is equivalent).

The question is how to extract information on the cohomology of $G$ (as a topological space) from the above.

For the complex cohomology there are no problems whatsoever. We only need $\mathfrak{g}$: restricting the complex formed by the left invariant forms to the unit of $G$ we get the standard cochain complex of $\mathfrak{g}$.

The next step would be the rational cohomology. One possible guess on how to get it would be to notice that $\mathfrak{g}$ is in fact defined over $\mathbf{Q}$. So one can find an algebra $\mathfrak{g}(\mathbf{Q})$ such that $\mathfrak{g}=\mathfrak{g}(\mathbf{Q})\otimes_\mathbf{Q}\mathbf{C}$. We can identify $H^{\bullet}(\mathfrak{g},\mathbf{C})=H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\otimes\mathbf{C}$ and so we get two rational vector subspaces in the complex cohomology of $G$. One is the image of of $H^{\bullet}(G,\mathbf{Q})$ and the other is the image of $H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})$ under $$H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\to H^{\bullet}(\mathfrak{g},\mathbf{C})\to H^\bullet(G,\mathbf{C})$$ where the last arrow is the comparison isomorphism mentioned above.

1). What, if any, is the relationship between these subspaces? More precisely, apriori the second subspace denends on the choice of $\mathfrak{g}(\mathbf{Q})$ (I don't see why it shouldn't, but if in fact it doesn't, I'd be very interested to know) and the question is if there is a $\mathfrak{g}(\mathbf{Q})$ such that the relationship between the above subspaces of $H^{\bullet}(G,\mathbf{C})$ is easy to describe.

Notice that this is somewhat similar to what happens when we compare the cohomology of the algebraic de Rham complex with the rational cohomology. Namely, suppose we have a smooth projective or affine algebraic variety defined over $\mathbf{Q}$; its algebraic de Rham cohomology (i.e. the (hyper)cohomology of the de Rham complex of sheaves) sits inside the complex cohomology, but this is not the same as the image of the topological rational cohomology. Roughly speaking, the difference between the two is measured by periods, e.g. as defined by Kontsevich and Zagier.

2). If question 1 has a reasonable answer, then what about the integral lattice in $H^{\bullet}(G,\mathbf{C})$? Again, a naive guess would be to take a $\mathfrak{g}(\mathbf{Z})$ such that $\mathfrak{g}(\mathbf{Z})\otimes\mathbf{C}=\mathfrak{g}$. At present, I'm not sure whether such an integral form exist for any reductive $G$ (and I'd be very interested to know that), but in any case it exists for $SL(n,\mathbf{C})$. By taking the standard complex of $\mathfrak{g}(\mathbf{Z})$ and extending the scalars we get a lattice in $H^{\bullet}(\mathfrak{g},\mathbf{C})\cong H^{\bullet}(G,\mathbf{C})$. Is there a choice of $\mathfrak{g}(\mathbf{Z})$ for which this lattice is related to the image of the integral cohomology in $H^\bullet(G,\mathbf{C})$ in some nice way?

3). If even question 2. has a reasonable answer, then what about the integral cohomology itself? Here, of course, the answer is interesting even up to isomorphism. A very naive guess would be to take an appropriate integral form $\mathfrak{g}(\mathbf{Z})$ as in question 2 and compute the integral cohomology of the resulting standard complex.

  • $\begingroup$ If you're working with Z-forms of split reductive groups, you should replace the Lie algebra with Kostant's universal enveloping algebra. $\endgroup$
    – S. Carnahan
    Commented Jan 14, 2010 at 18:41
  • $\begingroup$ Thanks, Scott. Replace to what purpose? $\endgroup$
    – algori
    Commented Jan 14, 2010 at 20:59
  • 1
    $\begingroup$ Even in the complex case, if G is noncompact, say SL(n,C), how are you obtaining its cohomology from the cochain complex for g? $\endgroup$
    – Tom Church
    Commented Feb 2, 2010 at 17:23
  • $\begingroup$ Tom -- right invariant forms on SL(n,C) are quasi-isomorphic to all forms. But I don't know how to show this directly. Instead, one can notice that the standard cochain complex of $sl(n,C)$ with $C$ coefficients is the same as the complexification of the standard complex of $su(n)$ with $R$ coeficients. The latter computes the complex cohomology of $SU(n)$ by averaging and the cohomology of $SL(n,C)$ is the same. $\endgroup$
    – algori
    Commented Feb 2, 2010 at 22:09

2 Answers 2


In question 1, it seems that the subspace $H^*(g(Q))$ in $H^*(g(C))$ indeed depends on the rational form of g. Consider the example of $sl(3)$ and two rational forms, $su(3,Q)$ and $sl(3,Q)$, and look at the fundamental classes in $H^8$ (i.e. the generators of the top exterior power). A basis of $su(3,Q)$ is $i(E_{11}-E_{22}), i(E_{22}-E_{33}), E_{12}-E_{21}, E_{13}-E_{31}, E_{23}-E_{32}, i(E_{12}+E_{21}), i(E_{13}+E_{31}), i(E_{23}+E_{32})$. Since we have 5 factors of i, the two fundamental classes differ by a factor of i (up to rational factors).

  • $\begingroup$ Thanks, Pavel! (Apparently, jsmath has an issue with asteriscs, so these have to be protected; I've also inserted the dollar signs.) $\endgroup$
    – algori
    Commented Feb 2, 2010 at 12:50
  • 1
    $\begingroup$ For what it's worth: using \ast for asterisks avoids this problem, as in $H^\ast(g(Q))$ $\endgroup$ Commented Jul 22, 2014 at 21:34

This is rather late, and not a full answer to your question, but may be interesting nonetheless.

Concerning point 1), I would like to propose the following natural refinement which avoids the problems mentioned in Pavel Etingof's answer. If you want to know the relation between cohomology of the group and the Lie algebra over $\mathbb{Q}$, you should work with $\mathbb{Q}$-forms of both. Take $G_{\mathbb{Q}}$ a form of $G$ defined over $\mathbb{Q}$, and take $\mathfrak{g}_{\mathbb{Q}}$ the associated Lie algebra (in the sense of algebraic groups). Then you want to compare the algebraic de Rham cohomology of $G_{\mathbb{Q}}$ with the Lie algebra cohomology of $\mathfrak{g}_{\mathbb{Q}}$. As Pavel Etingof pointed out, these $\mathbb{Q}$-vector subspaces of the cohomology of $G$ over $\mathbb{C}$ depend on the choice of form. Maybe there is an explicit formula which from a cocycle in $H^1_{\text{ét}}(\mathbb{Q},\operatorname{Aut}G)$ produces the factors relating the subspaces - the factor $i$ in Pavel Etingof's answer is most certainly related to defining $SU(3)$ in terms of the complex conjugation involution.

Concerning point 2), the same as in point 1) applies to $\mathbb{Z}$-forms: choose a $\mathbb{Z}$-form of $G$ and take its associated Lie algebra. You will most likely end up getting Kostant's $\mathbb{Z}$-forms mentioned in Scott Carnahan's comment.

In the special case $GL_n$ and cohomology in top degree, your question 2) and 3) are discussed in a preprint of Annette Huber and Wolfgang Soergel, arXiv version can be found here. The goal of that paper is to compare different integral structures in the top cohomology of $GL_n$: one integral structure comes from the natural $\mathbb{Z}$-form of $\mathfrak{gl}_n$, another from the suspensions of Chern classes in de Rham cohomology, and the third one from the dual of the fundamental class in singular cohomology. They work out the explicit comparison factors. The comparison between de Rham and singular cohomology involves the usual period $2\pi i$, and the comparison between de Rham cohomology and the subspace coming from the Lie group cohomology is in fact a rational factor (so no further periods).

The rational factor in the comparison between Lie algebra and de Rham cohomology that is worked out in the Huber-Soergel paper is $\prod_{j=1}^n(j-1)!$. It is interesting to note that these same factors appear when comparing homotopy and homology of $GL_n(\mathbb{C})$ - the Hurewicz map sends a generator of $\pi_{2j-1}GL_n(\mathbb{C})$ to some multiple of a generator of $H_{2j-1}(GL_n(\mathbb{C}),\mathbb{Z})$, and this multiple is $(j-1)!$. I am fairly convinced now that this is the way the subspaces are related in case $GL_n$ - the group cohomology classes in degree $2j-1$ should be the $(j-1)!$-th multiple of the generators of Lie algebra cohomology. One would hope that such a description applies to other Lie groups and that the rational factor can be explained by some Weyl group combinatorics in general, but there is some work to be done for that...

Concerning periods there is a philosophical reason why the comparison between de Rham cohomology and singular cohomology should only involve rational multiples of powers of $2\pi i$. If $G$ is a reductive group over $\mathbb{Q}$, we can view it as a variety and as such it has a mixed Tate motive; therefore, all periods should be rational multiples of powers of $2\pi i$.


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