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As in Algebra+Homotopy=Operad by Bruno Vallette, let $A$ with multiplication $\nu$ be a differential graded associative algebra equipped with degree +1 map $h$ and let $H$ be a chain complex such that there exist chain maps $i$ and $p$ such that

diagram from Vallette's paper

and I work in characteristic 2 to make everything easier. Define

$$\mu_2=p\circ\nu\circ(i\otimes i):H\otimes H\to H,$$

and in general,

mu_n.

is a degree $+(n-2)$ map, where $PBT_n$ means binary trees with $n$ nodes and Vallette's summand on the right is an example of a summand for $n=5.$ For $f\in\hom(H^{\otimes n},H),$ define $\partial f=d\circ f + f\circ d_{H^{\otimes n}}$; remember that we are in characteristic 2.

One way to visualize $\partial\mu_n$ is that the one term decorates the leaves with $d$'s, as $d$ is a derivation for the raw tensor product, and the other puts a $d$ at the root, which then propagates upwards, as $d$ is a derivation for $\nu.$

The Homotopy Transfer Theorem for Differential Graded Associative Algebras is that $H$ equipped with the $\mu_n$ is an $A_\infty$ algebra, which means precisely that

partial mu_n..

All images have been directly screencapped from Vallette's paper. He writes that it should be an "easy and pedagogical" exercise to prove this theorem, but I'm getting caught in the thicket even in this characteristic 2 case where there are far fewer $\pm$'s to keep track of. I was wondering if anyone could provide me with any insights as to how to proceed without trees popping up all over the place occluding the forest.

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2 Answers 2

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There is a systematic graphical notation that allows the tracking of signs, which I will mention at the end of this answer. But before doing so, let me outline the situation when $2 = 0$.

Note first that since $(A,\nu)$ is strictly associative, $\mu_0 = 0$ and $\mu_1 = d_H = d : H \to H$. By convention, if I have multilinear maps $f: H^{\otimes k}\to H$ and $g: H^{\otimes l}\to H$, then I will write $f\circ g : H^{\otimes(k+l -1)} \to H$ for: $$ (f\circ g) (x_1\otimes \cdots \otimes x_{k+l-1}) = \sum_{i=1}^k f\bigl(x_1\otimes \dots \otimes x_{i-1} \otimes g(x_i\otimes \dots \otimes x_{i+l-1}) \otimes x_{i+l} \otimes \cdots \otimes x_{k+l-1}\bigr) $$ This has a useful graphical notation, wherein the composition is the sum over all rooted planar trees with an $f$ at the bottom node and precisely one $g$ at one of the upper nodes.

Then axiom to be an $A_\infty$-algebra in characteristic $2$ is $ 0 = \sum_{j=0}^{n+1} \mu_j\circ \mu_{n+1-j} $, or, since $\mu_0 = 0$ and $\mu_1 = d$: $$ [d,\mu_n] = \sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j} $$ The right-hand side is a sum over all rooted planar trees with $n$ leaves and precisely two nodes, each of which has at least two branches from it.

To check this, the first thing to convince yourself is that the operator $[d,-] : f \mapsto d\circ f + f\circ d$ is a derivation of composition and tensor, so that to apply $[d,-]$ to some large diagram, you sum all diagrams you get by replacing one component of your original diagram by $[d,-]$ of it. Note also that $[d,-]$ comutes with (i.e. annihilates) $i$, $p$, and $\nu$. So when you work out $[d,\mu_n]$, you get a sum over diagrams that look like $\mu_n$ (i.e. planar rooted trees with $n$ leaves, each node has two branches, and interior edges labeled by $h$), except one of the interior edges has been replaced by $[d,h] = \mathrm{id}_A + ip$.

Now, it should be completely clear that the diagrams where the $h$ is replaced by an $ip$ are precisely the diagrams appearing in $\sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j}$. (If this is not clear, let me know, and I will try to make it clearer.)

Finally, we must dispense with the diagrams in which an $h$ is replaced by an $\mathrm{id}$. For any such diagram, consider contracting it along the offending $\mathrm{id}$ vertex, to produce a node with three branches. Except the resulting diagram with the trivalent vertex can be produced in two ways, corresponding to the two planar ways of blowing up a rooted node with three branches into two two-branch nodes. So, after sorting all of your offending diagrams into such pairs, you get a sum of diagrams that looks like a $\mu_n$-type sum, except one vertex is has three branches. What is this vertex labeled by? Why, $\nu \circ \nu$, of course, which is a sum of two terms. On the other hand, $\nu \circ \nu = 0$ by the associativity law for $(A,\nu)$.


In characteristic not equal to $2$, the exact same argument works, but you must find a good convention / notation for signs. The best notation that I know is as follows. It should, of course, already by understood that the solid "$H$" or "$A$" edges extend to "infinity" at the top and bottom of the page. You should additionally draw diagrams with some other color of edge (I usually used "dashed") that records the degrees of operators — so this "dashed" edge should carry an arrow denoting its direction. A vertex that raises homological degree by $n$ is required to receive $n$ dashed edges, and a vertex that lowers homological degree by $n$ is required to emit $n$ dashed edges. Free dashed edges are sent off to "infinity" at (say) the left-hand side of the page, and the order from top to bottom that the free dashed edges arrive is important. Just as you cannot add diagrams whose numbers of input and output "$H$" strands mismatch, you similarly must have the same sequence of dashed edges. (In categorical language, what I'm saying is that you only work with "global" elements of endomorphism spaces, which is to say actual morphisms in the category of homologically-graded abelian groups, but that you give yourself access to the objects of this category which are lines in degree $\pm 1$.)

Now whenever two edges cross, something happens with signs. The notation basically takes care of this, but if you ever insist on working with "homogeneous elements" (which is a bad habit — it's better to work more categorically) the convention is that as an element runs down the "wire" of a solid edge, when it passes through a dashed edge it remains unchanged if it is of even degree and changes sign if it is of odd degree. The notations that do matter are:

  1. A closed dashed circle can be removed for a factor of $-1$.
  2. A dashed crossing can be resolved for a factor of $-1$. (Since dashed edges are directed, any dashed crossing has a unique resolution.)

Resolving a crossing for a sign Loops, etc., cost signs

For example, the operator $d$ emits a dashed edge, and the operator $h$ receives a dashed edge. Thus an equation like "$\mathrm{id} - ip = dh + hd$" is nonsense: the left-hand side has no dashed edges running to infinity, whereas on the right-hand side the first summand emits an edge and then receives one, and the second summand does those in the opposite order. To sum the two terms on the right-hand side you have to at least get them into their edges-at-infinity into the same order, which you can do by adding a crossing (but remember that resolving that crossing changes a sign). To make the two sides agree, you get connect up the two dashed edges, and again you should think a moment about signs. The correct right-hand side to "$\mathrm{id} - ip = dh + hd$" is:

the commutator

Yes, the sign is correct. (Incidentally, I'm lifting these images from my thesis, which works out a slightly different question, and so the colors and numbers are not for this post.)

Anyway, I'll leave it as an exercise to write out diagrams for $A_\infty$ algebras in this notation, and to get all the signs right. (Hint: there should be no "weird" signs.) Part of the reason that I'll leave it as an exercise is that there's not really a unique correct answer: you make a sign convention, and work with it.

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If your tree had no $h$ in it, then when $d$ propagated upward, it would pass unchanged through $i$ and $p$, which are chain maps, and pass as a derivation through the product so you'd get a sum of terms which would cancel with the terms that have $d$ on top. Something similar will happen when you have $h$ on the edges, but you will get extra terms.

When you have $d$ below an $h$, that is equal to a sum of three terms. One is $d$ above an $h$, one is $ip$, and one is $\mathrm{Id}$. This is a local phenomenon that happens within the tree. We can take the term with $d$ above the $h$ and continue to allow the $d$ to propagate upward. At the end, the $d$s reach the top and cancel with the terms that have $d$ on top, so the sum turns out to be made up of two kinds of terms:

Terms where one $h$ has been replaced by $ip$

Terms where one $h$ has been replaced by $\mathrm{Id}$

Each $h$ is passed through precisely once by a $d$ as it propagates upward, so we have this sum over all internal edges of our tree.

The sum of terms of the first type is exactly the right side of your screen capture. The terms of the second type will cancel in pairs by the associativity of $\nu$. That is, there will be precisely two trees in the overall sum over $PBT_n$ that give rise to each term of the second type.

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