5
$\begingroup$

Consider two cochain DGA (differential graded algebras) named $A$ and $B$. By "coproduct" of two DGA I mean the category theory coproduct, not the coalgebra coproduct. It is defined in "A closed model structure for differential graded algebras" by J.F.Jardine. And cohomology of an algebra $A$ is defined by $H^{i + 1}(A)=\ker(d^{i+1})/\operatorname{Im}(d^{i})$. It is a natural DGA with zero differential. Then we can define the coproduct of two cohomology algebras. The question now becomes whether the following statement is true

$$H(A) \amalg H(B) \cong H(A \amalg B)$$

where the isomorphism is taken under $\mathbf{DGA}_k$.

I think it is true at least for DG algebras with the underlying structure of free algebras, with zero differential. I have written a proof. But I am still checking it.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ TeX note: \coprod is a ‘large’ operator, analogous to \prod. The binary operator, analogous to \times, is \amalg. Thus you want $H(A \amalg B)$ H(A \amalg B), not $H(A \coprod B)$ H(A \coprod B). I edited accordingly. $\endgroup$
    – LSpice
    Commented Mar 23 at 1:20

1 Answer 1

Reset to default
4
$\begingroup$

The coproduct in the category of non-unital dg algebras is maybe easier to think about. Indeed note that the two relations in Jardine's note only have to do with the units in the two algebras. The non-unital coproduct is given by $$ A \sqcup B = A \oplus B \oplus A\otimes B \oplus B \otimes A \oplus A \otimes B \otimes A \oplus B \otimes A \otimes B \oplus \dots $$ That is, we sum all possible tensor products in which the $A$'s and the $B$'s alternate. Multiplication is given by concatenating tensors and using multiplication in $A$ and $B$ whenever there are two adjacent factors of $A$ or $B$ in the result. The tensor product is the tensor product of $k$-modules. It follows from this formula that $H(A \sqcup B) \cong H(A)\sqcup H(B)$ e.g. when $k$ is a field, or more generally if both $A$ and $H(A)$ are flat over $k$.

It is interesting that the coproduct of associative algebras is so much simpler than the coproduct in commutative rings, i.e. the tensor product.

It's not immediately clear to me what happens in the presence of units, in full generality. "Most" dg algebras in nature are augmented, in which case you replace $A$ and $B$ by their augmentation ideals in the above formula, and add a unit.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ In the case where the underlying module of $A$ splits off the unit ($A = k \oplus \bar A$) and similarly for $B$, I believe that your formula for the coproduct does generalize to one that looks like $k$ plus the nonunital coproduct of $\bar A$ with $\bar B$. You can get many counterexamples in cases where it doesn't split off, e.g. when $A = k/J$ for some ideal $J$. $\endgroup$
    – Tyler Lawson
    Commented Mar 25 at 15:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .