what is the tensor product of two anti-commutative DG-algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:00:35Z http://mathoverflow.net/feeds/question/37199 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37199/what-is-the-tensor-product-of-two-anti-commutative-dg-algebras what is the tensor product of two anti-commutative DG-algebras? Heer 2010-08-30T22:59:19Z 2010-08-31T04:05:19Z <p>I mean what the multiplication law of the tensor product is.</p> http://mathoverflow.net/questions/37199/what-is-the-tensor-product-of-two-anti-commutative-dg-algebras/37213#37213 Answer by Theo Johnson-Freyd for what is the tensor product of two anti-commutative DG-algebras? Theo Johnson-Freyd 2010-08-31T04:05:19Z 2010-08-31T04:05:19Z <p>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.</p> <p>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 <code>$\mathfrak{d\!g}$</code>, a <em>differential graded vector space</em> is nothing more nor less than a <code>$\mathfrak{d\!g}$</code>-module (in $\mathbb Z\text{-Vect}$).</p> <p>Let <code>$\mathfrak{d\!g}\text{-mod}$</code> denote the category of representations of <code>$\mathfrak{d\!g}$</code>. 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 <em>differential graded algebra</em> is an algebra object in this category, and it is "anticommutative" in the classical sense if it is <em>commutative</em> in the categorical sense: the symmetric structure (the Koszul rule) determines for any two <code>$\mathfrak{d\!g}$</code>-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 <em>commutative</em> if <code>$m_A = m_A \circ \text{flip}_{A,A}$</code>.</p> <p>Given two algebras <code>$(A,m_A),(B,m_B)$</code> in any symmetric monoidal category, their tensor product is the algebra structure on $A\otimes B$ given by: <code>$$m_{A\otimes B} = (m_A \otimes m_B) \circ (\text{id}_A \otimes \text{flip}_{A,B} \otimes \text{id}_B) : A \otimes B \otimes A \otimes B \to A\otimes B$$</code> If $A,B$ are both commutative, so is $A\otimes B$.</p> <p>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.</p>