Metric tensor

1. Feb 14, 2010

barnflakes

I know that: $$g(a u\otimes v) = a g(u\otimes v)$$

where u and v are vectors and a is a constant, but what if a is a scalar field, is this rule also true?

ie. how do I interpret the expression:

$$g(f u\otimes u + v\otimes v)$$

where u and v are vector fields and f is a scalar field?

2. Feb 14, 2010

Fredrik

Staff Emeritus
Yes, it is. Tensor fields on a manifold M are always linear over $C^\infty(M)$, not just over $\mathbb R$. The only object that you'll encounter that isn't this nice is the connection, $(X,Y)\mapsto\nabla_XY$ which takes two vector fields to a vector field, and is $C^\infty(M)$ linear in the first variable (i.e. X), and only $\mathbb R$ linear in the second (i.e. Y).

I assume that your u and v are vector fields, not tangent vectors at a specific point p, and that g is the metric tensor field, not the metric tensor at p. Because it wouldn't make much sense to multiply a tangent vector at p with a scalar field.

3. Feb 14, 2010

barnflakes

Thank you Fredrik, so I'm interpreting the above expression to be $$f g(u \otimes u) + g( v \otimes v)$$ ?

Also, when calculating the inverse metric tensor, it's of the form $$g^{ab} \partial_{a} \otimes \partial_{b}$$ but are the g^(ab) elements of the inverse metric equal to the g_(ab) elements of the metric?

4. Feb 15, 2010

haushofer

No, not necessarily :) The metric

$$g^{ab} \partial_{a} \otimes \partial_{b}$$

can be seen as linear map from the space of vectors V to the space of dual vectors V*. These two spaces have the same dimension and are isomorphic. If you would write down the metric

$$g_{ab} dx^{a} \otimes dx^{b}$$

you would get a linear map from V* to V. So you suspect that first acting with the "contravariant metric" on a dual vector obtaining a vector and then acting with the "covariant metric" on this vector would give you the original dual vector again: $g^{ab} \partial_{a} \otimes \partial_{b}$ and $g_{ab} dx^{a} \otimes dx^{b} [/tex] are eachothers inverses. And the components of inverses don't have to be equal to eachother. This only happens if the metric components equal those of the Kronecker delta, the Minkowski metric or some other metric with [itex]\pm 1$ on the diagonal.

I hope Fredrik doesn't mind that I'm answering, maybe he has some comments on it :)

5. Feb 15, 2010

DrGreg

$$g^{ab}g_{bc} = \delta^a_c$$

6. Feb 16, 2010

haushofer

I see I interchanged "vector space V" and "vector space V*" with eachother.

7. Feb 16, 2010

George Jones

Staff Emeritus
I am curious: what do yo mean by $g(u\otimes v)$? At every event in spacetime, you are taking $g$ to be a map

$$g : V \otimes V \rightarrow ???$$

where $V$ is a tangent (vector) space. Is this really what you meant? This is possible, but usually $g$ is taken to to be a map

$$g : V \times V \rightarrow \mathbb{R},$$

with evlauation denoted $g \left( u , v \right)$. $V \otimes V$ and $V \times V$ are very different animals.

8. Feb 16, 2010

Altabeh

Some textbooks also use the set of smooth scaler functions $$C^{\infty}(M)$$ defined over the manifold M in place of $$\mathbb{R}$$.

AB

9. Feb 16, 2010

George Jones

Staff Emeritus
I know, but I wrote
i.e., I'm fixing an event $p$, in which case the map is into the set of real numbers.

10. Feb 16, 2010

Altabeh

Ah, my bad, I didn't pay attention carefully!

AB

11. Feb 16, 2010

Fredrik

Staff Emeritus
I didn't even notice that thing that George Jones brought up. I just saw a question about linearity and answered it. But I agree that the expression doesn't make sense as it stands.