How Differentials Compose

Next problem, Frankel’s 2.3(1), part(i):

Let F:M^n \rightarrow W^r and G:W^r \rightarrow V^s be smooth maps. Let x, y and z be local coordinates near p \in M, F(p) \in W , and G(F(p)) \in V , respectively. We may consider the composite map G \circ F : M \rightarrow V. Show, by using bases \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, and \frac{\partial}{\partial z}, that (G \circ F)_* = G_* \circ F_*.

Now, the differential of F, F_*, acts like this: F_* \left( \frac{\partial}{\partial x^j} \right) = \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \frac{\partial}{\partial F^i\left(x\right)}, where x = (x^1, x^2, \cdots , x^n) \in M^n and F^i(x) is the ith component of F(x). A similar definition holds for G_* and (G \circ F)_*.

The problem asks us to consider the differentials acting on the bases \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, and \frac{\partial}{\partial z}. However, it should be obvious that we need only prove that (G \circ F)_*\left(\frac{\partial}{\partial x}\right) = G_* \circ F_*\left(\frac{\partial}{\partial x}\right). The same proof would apply to the other two bases, and hence to the whole vector space that has those three bases as its basis.

Thus
G_* \circ F_* \left( \frac{\partial}{\partial x^j} \right)
= G_* \left( \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \frac{\partial}{\partial F^i\left(x\right)} \right)
= \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} G_* \left( \frac{\partial}{\partial F^i\left(x\right)} \right)
= \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \sum_k \frac{\partial G^k \left(F\left(x\right)\right)}{\partial F^i \left(x\right)} \frac{\partial}{\partial G^k\left(F\left(x\right)\right)}
= \sum_k \left(\sum_i \frac{\partial G^k \left(F\left(x\right)\right)}{\partial F^i \left(x\right)} \frac{\partial F^i\left(x\right)}{\partial x^j} \right) \frac{\partial}{\partial G^k\left(F\left(x\right)\right)}
= \sum_k \frac{\partial \left(G \circ F \right)^k \left(x\right)}{\partial x^j} \frac{\partial}{\partial G^k \left( F\left(x\right) \right)}
= (G \circ F)_* \left( \frac{\partial}{\partial x^j} \right)

Published in: on October 20, 2008 at 2:31 pm Leave a Comment

Failed, again.

So, I have not had time, or have been persuading myself that I don’t have time, to do what I originally intended to do with this blog. In lieu of making the (not inconsiderable) effort to explain technical stuff in ordinary language, I will occasionally post solutions to math/physics problems I have been doing. Perhaps with a little explanation thrown in.

Today’s problem is Problem 2.1(1) from Frankel’s The Geometry of Physics:
If v is a vector and \alpha is a covector, compute directly in coordinates that \sum_i a_i^V v^i_V = \sum_j a_j^U v_U^j . (Here V and U are the open sets associated with charts on the manifold.) What happens if w is another vector and one considers \sum_i v^i w^i ?

The first part is a matter of applying the formulae for coordinate transformations for vectors and covectors. For covectors, a_j^U = \sum_i a_i^V \frac{\partial x_V^i}{\partial x_U^j} . For vectors, v_V^i = \sum_j \frac{\partial x_V^i}{\partial x_U^j} v^j_U .

Starting from the left hand side of what we want to prove:
\sum_i a_i^U v^i_U = \sum_j \sum_i a_i^V \frac{\partial x_V^i}{\partial x_U^j} v_U^j = \sum_i a_i^V \sum_j \frac{\partial x_V^i}{\partial x_U^j} v^j_U = \sum_i a_i^V v_V^i

This shows that the scalar you get from letting a vector ‘eat’ a covector is coordinate independent. A covector is often defined as an entity that maps vectors to scalars. This is a coordinate independent definition — if you change coordinate systems, the same vector-covector combination should give you the same scalar. So it’s not at all surprising that \sum_i a_i^V v^i_V = \sum_j a_j^U v_U^j : computing that scalar in the coordinate system V should give us the same result as computing it in U.

The second part of the question asks if the ’scalar product’ of two vectors, \sum_i v^i w^i , is likewise independent of coordinate systems. Using the coordinate transformation formula for vectors, we get the following:
\sum_i v_U^i w_U^i = \sum_j \sum_i \left( \frac{\partial x_U^i}{\partial x_V^j} \right)^2 w^j_V v_V^j

So \sum_i v^i w^i is not coordinate independent. This, again, is not surprising. Unlike covectors, vectors are not defined as objects that take in other vectors as arguments and spit out a scalar. So we would not expect the coordinate independence of the above vector-covector combination to hold for vector-vector combinations.

Published in: on at 12:04 pm Leave a Comment