Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra (or the corresponding Lie group). For definiteness, I'll take $\mathfrak{g}$ to be of type $A_n$, that is, $\mathfrak{g} = \mathfrak{sl}_{n+1}(\mathbb{C})$, but my question applies to semisimple Lie algebras of arbitrary Lie type. Consider the Dynkin diagram for $\mathfrak{g}$. We can remove a node from the diagram to obtain a sub-diagram of type $A_{n-1}$. The sub-diagram corresponds to a copy of the Lie algebra $\mathfrak{g}':=\mathfrak{sl}_n(\mathbb{C})$ sitting inside $\mathfrak{g}$.

The structure of the Lie algebra cohomology rings $H^\bullet(\mathfrak{g},\mathbb{C})$ and $H^\bullet(\mathfrak{g}',\mathbb{C})$ are known, and are the same as the cohomology rings $H^\bullet(G,\mathbb{C})$ and $H^\bullet(G',\mathbb{C})$ for the corresponding complex Lie group. The computation of the cohomology rings is classical; for Lie algebras the computation is a result of Koszul.

In the specific case $\mathfrak{g} = \mathfrak{sl}_{n+1}(\mathbb{C})$, we have $H^\bullet(\mathfrak{g},\mathbb{C}) = \Lambda(x_3,x_5,\ldots,x_{2n+1})$, an exterior algebra on homogeneous generators of degrees $3,5,\ldots,2n+1$. Then $H^\bullet(\mathfrak{g}',\mathbb{C}) = \Lambda(x_3,x_5,\ldots,x_{2n-1})$. (For other Lie types, the cohomology ring is still an exterior algebra on homogeneous generators of certain odd degrees depending on the root system.)

The inclusion of Lie algebras $\mathfrak{g}' \rightarrow \mathfrak{g}$ gives rise to a corresponding restriction map in cohomology: $H^\bullet(\mathfrak{g},\mathbb{C}) \rightarrow H^\bullet(\mathfrak{g}',\mathbb{C})$.

Is the restriction map in cohomology map the "obvious'' map from $\Lambda(x_3,x_5,\ldots,x_{2n+1})$ to $\Lambda(x_3,x_5,\ldots,x_{2n-1})$, that is, the map that takes $x_i$ to $x_i$ for $1 \leq i \leq 2n-1$ and that takes $x_{2n+1}$ to zero? If so, can you provide a reference for this fact? For other Lie types, is the restriction map also the obvious map?

Edit: In response to the comments below, I think I should have phrased the question as follows:

Hopefully clarified version of question: Is there a choice of generators for the cohomology rings $H^\bullet(\mathfrak{g},\mathbb{C})$ and $H^\bullet(\mathfrak{g}',\mathbb{C})$ such that the restriction map in cohomology has the above described form?

I acknowledge that things could get messier for restriction maps in type $D_n$ and for types $E_6$, $E_7$, and $E_8$, because in those cases the degrees of the generators for the cohomolgoy ring aren't as well-behaved. But maybe something can still be said in general about the restriction map (e.g., when is it surjective?).

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    $\begingroup$ This question doesn't make sense as stated, since the homogeneous generators aren't canonical (although their degrees are). The isomorphism $H^*(\mathfrak{g})\simeq \Lambda(\mathfrak{g}^*)^{\mathfrak{g}}$ is functorial, so it's certainly true in the special case of the standard embedding of $\mathfrak{sl}_n$ into $\mathfrak{sl}_{n+1}$ for a suitable choice of generators (e.g. one arising from elementary symmetric functions generators of the symmetric invariants). $\endgroup$ Commented Aug 8, 2010 at 0:31
  • $\begingroup$ As Victor points out, only the degrees of homogeneous generators are canonical here. These have the form $2m_{i}+1$, where $d_{i} = m_{i}+1$ are degrees of basic polynomial invariants for the Weyl group and the $m_{i}$ are exponents determining eigenvalues of a Coxeter element. For most simple Lie algebras each $m_i$ occurs just once; the exception is type $D_{2n}$. So this complicates working with noncanonical choices of generators in your set-up. It does seem likely that restriction in cohomology respects at least the degrees, but I don't recall a reference. $\endgroup$ Commented Aug 8, 2010 at 10:56

1 Answer 1


The cohomology rings of Lie groups are not just rings but also Hopf algbras with a coproduct Δ coming from the group structure. Picking primitive (Δ(x) = x⊗1 + 1⊗x) homogeneous generators gets close to giving a canonical set of generators up to scalars (though there is still some ambiguity for Dn which can have 2 generators of the same degree).

Homomorphisms of groups preserve the coproduct, so they will more or less preserve the primitive generators.

  • $\begingroup$ Okay, this seems to answer the question once one establishes that the restriction map doesn't just map the primitive generators to zero. $\endgroup$ Commented Aug 9, 2010 at 15:23

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