# what is the tensor product of two anti-commutative DG-algebras?

I mean what the multiplication law of the tensor product is.

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Presumably, you want the product in the category of graded algebras, so there is a Koszul sign coming in: $a\otimes b\cdot c\otimes d=(-1)^{|b|\,|c|}ac\otimes bd$ for homogeneous elements $a$, $b$, $c$ and $d$ in the factors, with $|a|$ denoting the degree of $a$. In any case, you should read the FAQ for suggestings of other sites where you'll have a much better fit for your question! – Mariano Suárez-Alvarez Aug 30 '10 at 23:11
I'm usually one to quickly pronounce questions "inappropriate for MO". In this case, I think a question like this is perfectly reasonable --- I've certainly asked trivial questions when trying to learn a new topic in mathematics. I would prefer if this question were better written: the title should begin with a capital letter, and the body should include more background, etc. mathoverflow.net/howtoask (linked at the top of the page) has more instructions on how this question should be revised. – Theo Johnson-Freyd Aug 31 '10 at 3:43
I think that this question is mis-stated. I think that it wants to ask for the tensor product of two graded commutative dgas, certainly that's what's being answered. It is in fact asking a different question. An anti-commutative dg-algebra satisfies a.b = (-1)^{|a||b|+1}b.a. The associativity and the resulting pentagon diagram means that the product of 4 elements is 0. I guess that the tensor product of two such algebras is naturally a graded commutative algebra but still with the product of 4 elements being 0. – James Griffin Aug 31 '10 at 9:41
@James Griffin: Yes, I think you're correct in your read. I guess the correct thing to do is to take differential graded vector spaces, and consider Z/2-graded objects within those, with another Koszul rule. That's of course not always anticommutative - only the odd terms anticommute - , but I have a hard time imaging anything I'd want to call an (associative, unital) algebra that's completely anticommutative, as I think that the identity element should commute with everything. Alternately, we could add the word "lie", and think about DGLAs. But there is not a tensor product of Lie algebras. – Theo Johnson-Freyd Aug 31 '10 at 15:32

The correct setting for differential graded vectors spaces is as follows. Recall first the category of $\mathbb Z$-graded vector spaces. As a category this consists of functors from the set $\mathbb Z$ (thought of as a category with no morphisms) to $\operatorname{Vect}$, i.e. objects consist are sequences of vector spaces, and morphisms are sequences of linear maps. (There are variations: one can insist that the vector spaces be trivial except for finitely many of them, for example, and that the non-trivial vector spaces be finite-dimensional.) As a monoidal category, the tensor structure adds degree. Only when you introduce the braiding does the category become interesting: we will take the "Koszul" braiding, so that classically odd-degree terms "anticommute". This braiding is symmetric.

Within the symmetric monoidal category $\mathbb Z\text{-Vect}$ of $\mathbb Z$-graded vector spaces there is a special Lie algebra, which is the unique (necessarily commutative) Lie algebra structure on the $\mathbb Z$-graded vector space with one dimension in degree $1$ and all other degrees trivial. (The only bracket is the $0$ one because, by construction, the bracket must add degree, and so must land in the degree-two part, which is zero-dimensional.) Suggestively calling this Lie algebra $\mathfrak{d\!g}$, a differential graded vector space is nothing more nor less than a $\mathfrak{d\!g}$-module (in $\mathbb Z\text{-Vect}$).

Let $\mathfrak{d\!g}\text{-mod}$ denote the category of representations of $\mathfrak{d\!g}$. It is a symmetric monoidal category on account of it being the representation theory of a Lie algebra: the symmetric monoidal structure is inherited from $\mathbb Z\text{-rep}$, so in particular there is the Koszul rule. A differential graded algebra is an algebra object in this category, and it is "anticommutative" in the classical sense if it is commutative in the categorical sense: the symmetric structure (the Koszul rule) determines for any two $\mathfrak{d\!g}$-modules $A,B$ a canonical isomorphism $\text{flip}_{A,B}: A\otimes B \to B\otimes A$, and an algebra $m_A: A\otimes A \to A$ is commutative if $m_A = m_A \circ \text{flip}_{A,A}$.

Given two algebras $(A,m_A),(B,m_B)$ in any symmetric monoidal category, their tensor product is the algebra structure on $A\otimes B$ given by: $$m_{A\otimes B} = (m_A \otimes m_B) \circ (\text{id}_A \otimes \text{flip}_{A,B} \otimes \text{id}_B) : A \otimes B \otimes A \otimes B \to A\otimes B$$ If $A,B$ are both commutative, so is $A\otimes B$.

This categorical mumbo-jumbo exactly recovers the multiplication that you are looking for. I hope also that it illustrates that it is very naturally part of a larger story, and does not come out of the blue.

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This type of answer is exactly why this type of question should stay open, at least for a little while. In particular, this is a very nice conceptual answer. The second best explanation of the Koszul sign rule I can remember hearing is the uninspiring "because it works." (Of course, this is my fault for not asking and/or for forgetting.) – Peter Samuelson Aug 31 '10 at 5:19
@Peter Samuelson: Thanks. Then again, I'm on record for siding with "good answers are not a sufficient condition for good questions". So my current opinion is that the original question should stay open only with some revisions. – Theo Johnson-Freyd Aug 31 '10 at 15:28
Yeah, I think I'd agree, because putting no effort into writing a question and expecting nice answers is unreasonable. Perhaps if this is closed it should be edited afterwards so that if someone later has the question that you answered it would be easier to find. – Peter Samuelson Sep 1 '10 at 6:49
I disagree with the implicit assertion that there is a unique correct setting for dg vector spaces. For example, one could just as easily replace the graded vector spaces step with an introduction to the category of super vector spaces, and take the category of representations of the algebraic supergroup $\mathbb{G}_a^{odd} \rtimes_{1} \mathbb{G}_m$. – S. Carnahan Sep 1 '10 at 12:24