# Cochain of Algebra

+ 5 like - 0 dislike
424 views

Hi. Can someone help me understand what the cochain of an algebra is? I cannot find something helpful in Google nor in my group theory or mathematics books (e.g. Nakahara).

Thanks a lot.

I guess it refers to (co)homological algebra, you can find related introductions in algebraic topology textbooks, e.g. Bredon, Hatcher etc.

It would help if you would add some context.  For example, what do you mean by algebra?  Do you mean an associative algebra, a Lie algebra, a Leibniz algebra, a Jordan algebra,...?  Each kind of algebra will have an associated cohomology theory and a cochain will be an element in a complex which computes this cohomology, but the precise definitions require knowing which kind of algebra you have in mind.

Thanks both for your answers. Well, I am working on non-commutativity and non-associativity. I am looking into Lie algebra deformations caused by, e.g., a magnetic monopoles in the presence of a background spherically symmetric magnetic field. From what I read, the presence of the monopoles destroy the spherical symmetry of the problem. Then the authors mention that in such cases there is an apparent Lie algebra deformation and then they give info about the cochains.

Not sure if it gives some context. Please keep in mind I am a physicist.

Thanks a lot.

+ 5 like - 0 dislike

Let $\mathfrak{g}$ be a Lie algebra and let $[-,-]:\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ be the Lie bracket.  By a (Lie algebra) deformation of $\mathfrak{g}$ one usually means a (typically, analytic) one-parameter family of Lie brackets $[-,-]_t$ on the same underlying vector space $\mathfrak{g}$, agreeing with the original Lie bracket at $t=0$.  A deformation is trivial if $[-,-]_t$ is obtained from $[-,-]$ by a one-parameter group $g(t)$ of linear transformations of $\mathfrak{g}$.  The derivative at $t=0$ of $[-,-]_t$ defines a skew-symmetric bilinear map $\phi: \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ which is a cocycle in the Chevalley-Eilenberg complex $C^2(\mathfrak{g};\mathfrak{g})$, whereas if the deformation is trivial, this cocycle is actually the coboundary of some cochain in $C^1(\mathfrak{g};\mathfrak{g})$.  Therefore, infinitesimal nontrivial deformations are classified by the Chevalley-Eilenberg cohomology group $H^2(\mathfrak{g};\mathfrak{g})$.  You can work out easily what the cocycle condition by taking the derivative at $t=0$ of the Jacobi identity for the deformed bracket $[-,-]_t$.

You can work in a basis $X_a$ for $\mathfrak{g}$ and then $[-,-]_t$ is defined by structure "constants" (which are now a function of $t$!): $C_{ab}{}^c(t)$ and the cocycle is the derivative at $t=0$ of these structure constants.

Given a nontrivial infinitesimal deformation, one is typically interested in whether it will integrate to a one-parameter family of deformations.  There is an infinite number of obstructions to integrability: each defined provided the previous one is overcome, which are classes in $H^3(\mathfrak{g};\mathfrak{g})$.

So, for example, if you can calculate $H^2(\mathfrak{g};\mathfrak{g})$ and $H^3(\mathfrak{g};\mathfrak{g})$ you are part of the way to understanding the deformation theory of $\mathfrak{g}$.  Computing Chevalley-Eilenberg cohomology is simply a question of solving linear equations, so it is doable in principle and particularly amenable to computer calculations.  But there are several results of a general nature which can save you lots of computation.  For example, if $\mathfrak{g}$ is semisimple, then $H^2(\mathfrak{g};\mathfrak{g}) = 0$, so that such algebras are (infinitesimally) rigid.  This means that any deformation is really just a $t$-dependent change of basis.

In your comment you mention rotational symmetry.  The rotation algebra in dimension $d>2$ is semisimple, so in fact it is rigid.  Any infinitesimal deformation you find is certainly trivial.

You can read about this in the original paper: C. Chevalley, S. Eilenberg, Cohomology theory of Lie groups and Lie algebras, Trans. Amer. Math. Soc. 63, (1948). 85–124.  It is very well written and not very abstract at all.  You may also benefit from the references in the answers to this MathOverflow question.

answered Jul 2, 2014 by (2,315 points)

Wow, this is very well written, and you have provided me with some really useful insight. Thank you very  much sir. If you, additionally, have to recommend a textbook I would appreciate it. Thanks a lot.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.