Interpreting g(f u\otimes u + v\otimes v) with Scalar Field f

[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?

 

[Fredrik]

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.

 

[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?

 

[haushofer]

No, not necessarily :) The metric

<br /> g^{ab} \partial_{a} \otimes \partial_{b} <br />

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

<br /> g_{ab} dx^{a} \otimes dx^{b} <br />

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 each others inverses. And the components of inverses don&#039;t have to be equal to each other. This only happens if the metric components equal those of the Kronecker delta, the Minkowski metric or some other metric with \pm 1 on the diagonal.<br /> <br /> I hope Fredrik doesn&#039;t mind that I&#039;m answering, maybe he has some comments on it :)

 

[DrGreg]

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

 

[haushofer]

I see I interchanged "vector space V" and "vector space V*" with each other.

 

[George Jones]

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

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

<br /> g : V \otimes V \rightarrow ?<br />

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

<br /> g : V \times V \rightarrow \mathbb{R},<br />

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

 

[Altabeh]

<br /> g : V \times V \rightarrow \mathbb{R},<br />

with evlauation denoted g \left( u , v \right).

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

AB

 

[George Jones]

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

AB

I know, but I wrote

At every event in spacetime,

i.e., I'm fixing an event p, in which case the map is into the set of real numbers.

 

[Altabeh]

i.e., I'm fixing an event p, in which case the map is into the set of real numbers.

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

AB

 

[Fredrik]

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.