Failed, again. « Explicating Thingummyhoos

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.