It is well-known that the space of all derivations of algebra of smooth functions on a manifold is its space of sections (vector fields on underlying manifold). But, I wonder to know why finding the space of all algebra derivations is an important problem? What aspects of its underlying algebra is this space capable to show ?
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It lets you set up a bijection between vector fields and derivations, so that you can alternate between the two freely – Zach Effman Aug 02 '15 at 14:41
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3@ZachEffman Which leads one to ask: why do we care about alternating between these? – Aug 02 '15 at 15:14
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To put it shortly, the interpretation of vector fields as derivations on the algebra of smooth functions leads to a definition of the Lie bracket of vector fields which is manifestly independent of coordinates. (So it is not the question of understanding the algebra of smooth functions, but rather that the interpretation as derivations gives you new ways to work with vector fields.)
When thinking about analysis on manifolds, there is the basic quest to find operations, which are independent of the choice of local coordinates, since these are the basic natural operations that you can carry out on any manifold. One way to find such operations is to write out coordinate expressions and verify independence of the choice of coordinates by a computation. While this is certainly a possible approach some people (including myself) view this as rather unsatisfactory. In fact, there are only very few operations that can be defined independently of coordinates, with the exterior derivative of differential forms, the Lie bracket of vector fields, and the Lie derivative of tensor fields forming the basic example. Hence it is natural to ask for a conceptual way of understanding why these operations are independent of coordinates (and why there are no others, but that's a different story). The most convincing way is that you give a description of the operation, which does not use coordinates at all.
Already the action of vector fields on smooth functions is an example of these ideas. A vector field $\xi$ associates to each point a tangent vector in that point, so given a smooth function $f$ so, you can associate to each $x$ the derivative of $f$ in the point $x$ in direction $\xi(x)$. This gives a coordinate free association of a linear map $C^\infty (M,\mathbb R)\to C^\infty (M,\mathbb R)$ to a vector field $\xi$, which we write as $f\mapsto \xi\cdot f$. It is easy to see that this is always a derivation, and a more complicated proof shows that any derivation is of this form. But a two line computation shows that the commutator of two derivations of an algebra is again a derivation. Hence given two vector fields $\xi$ and $\eta$ on $M$, there is a unique vector field $[\xi,\eta]$ on $M$ such that $[\xi,\eta]\cdot f=\xi\cdot (\eta\cdot f)-\eta\cdot (\xi\cdot f)$ and all important properties of the Lie bracket (as well as its expression in local coordinates) can be easily deduces from this definition, which is completely coordinate-free.
In fact, you can then use the Lie bracket to give coordinate-free descriptions of the exterior derivative and the Lie derivative of tensor fields.
Andreas Cap
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1Thank you dear Andreas, i found another answer to my question. The space of all derivations of an algebra has lie algebra structure and in fact, it is the lie algebra of lie group of automorphisems of underlying algebra. – user258413 Aug 03 '15 at 11:26
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This is indeed closely related. To see that derivations form a Lie algebra, you only need that the commutator of two derivations is again a derivation. In the context of vector fields, the relation to a group is given by the group of diffeomorphisms of the manifold in question (which also happens to coincide with the automorphism group of the associative algebra $C^\infty(M,\mathbb R)$). But the Lie group / Lie algebra issues are rather involved in infinite dimensions. – Andreas Cap Aug 03 '15 at 11:38