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Within the book An introduction to homological algebra by Weibel, I am trying to prove the following isomorphism, but I am not sure this is true. But I really want to know how to prove or disprove it.

Notation: Let $\mathfrak{g}$ be a finite dimensional Lie algebra over a field. Let $\text{H}^i_{\text{Lie}}(\mathfrak{g},A)$ be the i-th cohomology group of $\mathfrak{g} $ with the coefficients in the $\mathfrak{g}$-module $A$.

Page 226: Let $M,N$ be $\mathfrak{g}$-modules. Then we have the following natural isomorphism of $\delta$ functors:

$$ \text{Ext}^\ast_{U \mathfrak{g}}(M,N) \cong \text{H}^\ast_{\text{Lie}}(\mathfrak{g}, \text{Hom}_{k}(M,N)) $$

where $\ast$ is a non-negative integer.

  • I am thinking about the following possible way to prove this problem:

Both side are $\delta$-functor of $N$. Therefore it remains to show that the $\text{H}^\ast_{\text{Lie}}(\mathfrak{g}, \text{Hom}_{k}(M,-))$ is a universal $\delta$-functor. For this, we have to show that it vanishes on injectives (and so effacable). I think that it is possible to show that if $Q$ is a injective $\mathfrak{g}$-module then $\text{Hom}_{k}(M,Q)$ is also a injective $\mathfrak{g}$-module. But I can not prove this.

Thanks very much!

Najib Idrissi
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Ken Park
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  • Here $\text{Hom}_{k}(M,N) \cong M* \otimes N $ as $\mathfrak{g}$-modules. – Ken Park Jul 03 '13 at 03:35
  • Since $$\text{H}^\ast_{\text{Lie}}(\mathfrak{g}, \text{Hom}{k}(M,N)) = \text{Ext}^\ast{U(\mathfrak g)}(k, \text{Hom}{k}(M,N)),$$ I guess that it would be sufficient to check that $$\mathrm{Hom}{U(\mathfrak g)}(k, \mathrm{Hom}k(M, N)) \cong \mathrm{Hom}{U(\mathfrak g)}(k \otimes_k M, N) = \mathrm{Hom}{U(\mathfrak g)}(M,N),$$ using the fact that this identity holds when $\mathrm{Hom}{U(\mathfrak g)}(-,-)$ is replaced by $\mathrm{Hom}_k(-,-)$. – Watson May 13 '18 at 13:22

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