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Product rule

Xedi

Posted: Submitted by Xedi on 22 April 2008 - 12:45pm.

Joined: 2007-10-03
Posts: 375

This might be a silly question, but what types of operations satisfy the product rule ?
Obviously differentiation of functions does with

$d(uv) = u\,dv + du\,v$

Same with grad (which is just a consequence)

$\nabla(\phi\psi)=\phi\nabla(\psi)+\nabla(\phi)\psi$

etc.
Is it something exclusive to differentiation (as it seems somewhat to be as differentiation in some contexts is defined by the product rule) ?

Newtonswig

Posted: 22 April 2008 - 1:21pm

Joined: 2006-10-01
Posts: 427

You've happened upon something rather sexy here Sam. On a manifold differentiation is a tricky business, or at the least requires some fiddling: differentiation is a linear business, and only likes linear things, to make it make sense on a manifold we must associate to each point a linear space from which to do linear business.

One way is to define it in terms of the charts (the maps $\phi:M^n \rightarrow \mathbb{R}^n$ which give your manifold that exciting locally flat structure that defines them) and use the chain rule to see what happens to the flat bit around $\phi(x)$ .

Another way is to take the space of smooth functions $M\rightarrow \mathbb{R}$ denoted $C^{\infty}(M)$ , which form an algebra over $\mathbb{R}$ (that is, real vector space with multiplication), and then take its space of derivations.

Given an algebra $\mathfrak{A}$ over a field F, its space of derivations is precisely the set of F-linear maps $D:\mathfrak{A}\rightarrow \mathfrak{A}$ which satisfy the 'product rule' called the Liebniz rule by differential geometers.

What is interesting is that these two definitions are entirely equivalent (that the first implies the second is obvious- tangent space = tangent vectors = stuff to diffferentiate w.r.t. The second is less so- essentially a consequence of the structure of derivations over $C^{\infty}(\mathbb{R})$ from which a local isomorphism is induced by charts). Using this definition can open up worlds of abstraction, and indeed can be extended if the codomain of the algebra is not simply $\mathbb{R}$ , giving the notion of connection...

But yeah, derivations are ace.

Xedi

Posted: 24 April 2008 - 12:57pm

Joined: 2007-10-03
Posts: 375

Thanks, I've read up a bit and now understand your post :p