# Thread: Tensor bundle

1. ## Tensor bundle

Hello,

i try to show that one can define a metric on all tensor bundles with the riemannian metric.

I have really no idea, how i can show this. Here are the definition, which i know:
A Riemannian Metric on a Manifold M is a map f which assigns to each point p this:
$\displaystyle f(p):T_{p}M\;\times\;T_{p}M->\mathbb{R}$ which assigns to each point p a positive definite symmetric bilinearform.

And the tensor bundles are of the form:
$\displaystyle T_{r,s}(M):=\bigsqcup_{m\in M}(T_{m}M)_{r,s}$
with $\displaystyle (T_{m}M)_{r,s}:=T_{m}M\otimes...\otimes T_{m}M\otimes (T_{m}M)^{*}\otimes...\otimes (T_{m}M)^{*}$

Ok, meanwhile i'm not sure what the word "metric" mean. A metric on all tensor bundles. Is it a metric in the usual sense, that is a metric space?
Or is it a metric like the riemannian metric is. I.e. a positive definite symmetric bilinearform ??

Regards

2. Consider a vector space V with an inner product $\displaystyle \langle , \rangle$ defined. The inner product induces a canonical isomorphism $\displaystyle \phi$ between V and V*: $\displaystyle (\phi(v))(w) = \langle v, w \rangle$. So we can identify V and V* by this isomorphism. Then we can consider only tensor products $\displaystyle V_{r,0}$.
Define an inner product on $\displaystyle V_{r,0}$ as follows:
$\displaystyle \langle v_1 \otimes v_2 \otimes ... \otimes v_r, w_1 \otimes w_2 \otimes ... \otimes w_r \rangle = \langle v_1, w_1 \rangle \langle v_2, w_2 \rangle ... \langle v_r, w_r \rangle$

3. Excuse me this was a bad mistake of me! In europe it is very early in the morning! ;-)

Regards

4. why did you say that I was considering only for even ranks?

5. Excuse me! You are absolutely right. I did a mistake. I have forgotten that the inner product takes two elements of the tensor bundle therefor you are right!

Now i have to show that this construction is indeed continuous. Why do we have this property?

Regards

6. Since the induced inner product is (smoothly) expressible by the original one g, and g is smooth, we're done.
You can write down the exact expression to be sure.

7. Thank you again!

I think they are continuous, since our g is the product of the original one g (which is smooth).
So now we have defined a smooth inner product on each Tensor space of the form (r,s).(with r,s fixed).

What does this construction mean? What can i do with it? Or better: What do we need it for?

Do you know perhaps what this meaning of this construction is?

Regards