# Connections and tensor fields

1. Feb 23, 2015

### CAF123

1. The problem statement, all variables and given/known data
Let $T$ be a $(1, 1)$ tensor field, $\lambda$ a covector field and $X, Y$ vector fields. We may define $\nabla_X T$ by requiring the ‘inner’ Leibniz rule, $$\nabla_X[T(\lambda, Y )] = (\nabla_XT)(\lambda, Y ) + T(\nabla_X \lambda, Y ) + T(\lambda, \nabla_X Y ) .$$

(a) Prove that $\nabla_XT$ defines a $(1, 1)$ tensor field.
(b) Prove that the components of the $(1, 2)$ tensor $\nabla T$ are, $$\nabla_cT^{a}_{\,\,\,b} = e_c(T^{a}_{b}) + \Gamma^{a}_{\,\,dc}T^{d}_{b} − \Gamma^{d}_{ bc}T^{a}_{d}$$
(c) Deduce that the Kronecker delta tensor is covariantly constant $\nabla \delta = 0.$
2. Relevant equations

Tensor fields and linearity

3. The attempt at a solution

Really just need to check what I am going to do is correct. For a), $\nabla_X T$ defines a (1,1) tensor if it is linear in the arguments $X$ and $Y$? Linear in $X$ by definition of a connection and for $Y$;

$$\nabla_X (T (\lambda, fY)) = (\nabla_X T)(\lambda fY) + T(\nabla_X \lambda, fY) + T(\lambda, \nabla_X (fY))$$ which can be written using the axioms of a connection $$f(\nabla_X T)(\lambda, Y) + fT(\nabla_X \lambda, Y) + fT(\lambda, f \nabla_X Y) + T(\lambda, X(f)Y)$$ It doesn't seem to be linear in Y at all?

Thanks!

2. Feb 23, 2015

### stevendaryl

Staff Emeritus
Well, you started with the equation:

$\nabla_X (T(\lambda, Y)) = (\nabla_X T)(\lambda, Y) + T(\nabla_X \lambda, Y) + T(\lambda, \nabla_X Y)$

You can rewrite this to get:

$(\nabla_X T)(\lambda, Y) = \nabla_X (T(\lambda, Y)) - T(\nabla_X \lambda, Y) - T(\lambda, \nabla_X Y)$

So it's the combination of all three terms on the right-hand side that must be linear in $\lambda$ and $Y$. The individual terms don't have to be linear. To prove it's linear in $Y$, replace $Y$ by $f Y$:

$(\nabla_X T)(\lambda, f Y) = \nabla_X (T(\lambda, f Y)) - T(\nabla_X \lambda, f Y) - T(\lambda, \nabla_X (f Y))$

Then you have to prove that the right-hand side is equal to $f (\nabla_X (T(\lambda, Y)) - T(\nabla_X \lambda, Y) - T(\lambda, \nabla_X Y))$

To prove this, you have to use the linearity of $T$ and the Leibniz rule to rewrite the right-hand side so that hopefully you can get $f$ on the outside. What that means is that derivatives of $f$ have to cancel.

3. Feb 24, 2015

### CAF123

Thanks! Ok so I proved the linearity in $Y$, can I use the same method to find the linearity in $\lambda$? So let $\lambda \rightarrow f \lambda$?

For part b), I am not sure how much detail is required for an answer. The definition I have in my notes is that 'Given an $(r,s)$ tensor field $T$, the covariant derivative $\nabla T$ is an $(r,s+1)$ tensor with components $$\nabla_c T^{a_1 .... a_r}_{\,\,\,\,\,\,b_1 ....b_s} = e_c T^{a_1....a_r}_{\,\,\,\,b_1...b_s} + \Gamma^{a_1}_{dc} T^{d.....a_r}_{\,\,\,\,b_1....b_s} + .... + \Gamma^{a_r}_{dc}T^{a_1 ....d}_{\,\,\,\,b_1 ..... b_s} - \Gamma^d_{b_1 c} T^{a_1....a_r}_{\,\,\,\,d....b_s} - .... - \Gamma^{d}_{b_s c} T^{a_1...a_r}_{\,\,\,\,\,b_1....d}$$ So applying this gives the result immediately, as far as I can see we are evaluating all the possible contractions of the covariant and contravariant components of the tensor. Do you think this is fine? And is this result a definition or does it follow from something? Thanks!