Let $A$ be a $\mathbb{N}$-graded, locally finite $\mathbb{k}$-algebra, $\mathbb{k}$ being a field, $A = \bigoplus_{n \geq 0} A^n$, each $A^i$ being finitely-dimensional as a $\mathbb{k}$ vector space. Assume also that $A$ is connected, i.e., $A^0 = \mathbb{k}$.

Let $A^\ast$ denote its graded dual, i.e., $A^\ast = \bigoplus_{n \geq 0} \mathrm{Hom}_{\mathbb{k}}(A^n, \mathbb{k})$, which is a connected, graded $\mathbb{k}$-coalgebra.

Starting from some computations involving the (co)bar complexes for Hochschild (co)homology, I came up with the following question:

What is the connection between $\mathrm{Tor}_\bullet^A(\mathbb{k},\mathbb{k})$ and $\mathrm{Ext}^\bullet_{A^\ast} (\mathbb{k},\mathbb{k})$?

Computing them using the normalized (co)bar complexes, one obtains, in both cases, bigraded structures, $\mathrm{Ext}^\bullet_{A^\ast} (\mathbb{k}, \mathbb{k})$ being the Yoneda algebra of $A^\ast$ and $\mathrm{Tor}_\bullet^A(\mathbb{k}, \mathbb{k})$ being a coalgebra, obtained in a “dual” way as for the Yoneda algebra.

It would be great to have a duality between these structures, but I don't know how to approach an attempt of proof.

Thank you.

  • $\begingroup$ After reading some more about this subject and trying to simplify the question as much as possible, I'm starting to think that one could use Universal Coefficient Theorem for cohomology. How exactly, I'm still wondering... $\endgroup$ Feb 4, 2014 at 17:36

1 Answer 1


If $k$ is a field as you say it is, it is not difficult to see using the standard tensor-hom adjunction that $\mathrm{Ext}^*_R(k, k) \cong (\mathrm{Tor}_*^R(k, k))^*$ as $k$-graded modules, where $R$ is a $k$-algebra. I've learned this from these $\mathrm{Tor}$-$\mathrm{Ext}$ notes by May.

Here's the relevant passage, which contains other comments which might be useful to you. It's at the very end of the note:

Remember that when $R$ is commutative the $\mathrm{Ext}$ groups and all maps in sight between them take values in the category of $R$-modules. More generally, if $R$ is an algebra over a commutative ring $k$, then the $\mathrm{Ext}$ groups and all maps in sight between them take values in the category of $k$-modules. When $R$ is commutative and augmented over $k$, we have seen that $\mathrm{Tor}^R_*(k, k)$ is a graded $k$-algebra. We now see that $\mathrm{Ext}_R^*(k, k)$ is also a graded $k$-algebra, via the Yoneda product. If $k$ is a field and $X$ is an $R$-free resolution of $k$, then we have $$ \mathrm{Hom}_R(X, k) ≅ \mathrm{Hom}_k(k ⊗_R X, k) $$ as $k$-chain complexes. Writing $M^* = \mathrm{Hom}_k(M, k)$ for the vector space dual of $M$, this implies that $$ \mathrm{Ext}_R^*(k, k) ≅ (\mathrm{Tor}^R_*(k, k))^* \,. $$ Therefore $\mathrm{Ext}_R^*(k, k)$ has both an algebra structure and the dual of an algebra structure, which is called a coalgebra structure. We shall return to consideration of such structures later, when we shall talk about bialgebras and Hopf algebras.

For now, we sum up by saying that for any $k$-algebra $R$, $\mathrm{Ext}_R^*(k, k)$ is a $k$-algebra under the Yoneda product. It is not necessarily commutative even when $R$ is commutative, but then $\mathrm{Ext}_R^*(k, k)$ is a Hopf algebra.

We shall later return to this point and show that if $R$ is a Hopf algebra over $k$, then $\mathrm{Ext}_R^*(k, k)$ is a commutative $k$-algebra, but not necessarily a Hopf algebra. As we shall see, when specialized to group algebras this is closely related to the fact that the cohomology of a space with coefficients in $k$ is a commutative algebra.

(Original scan)


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