# Metric Tensor

1. Jan 24, 2010

### latentcorpse

i'm writing an essay on General Relativity and am giving a definition of the metric. does this seem ok:

A $\emph{metric}$ is a nondegenerate symmetric bilinear form, $g$, on each tangent space to the manifold, $M$. $g$ has an invertible matrix of components, $g_{ab}$, satisfying $g_{ab}=g_{ba}$. The metric is a tensor which acts on a pair of vectors in each tangent space to produce a real number. That is:
$g:V_p \times V_p \rightarrow \mathbb{R}; (X,Y) \mapsto g_{ab}X^aY^b \quad \text{where } V_p \text{ is the tangent space at } p \in M$

feel free to add/change if its wrong or not precise enough. i'm looking for an accurate defn. thanks.

i am a bit confused as to whether X is in M or V_p. any advice?

Last edited: Jan 24, 2010
2. Jan 24, 2010

### Altabeh

Your definition sounds sort of incomplete as it lacks this important property of metric tensor that it varies smoothly throughout the manifold M with p, but let me give you a better one:

A metric tensor, denoted by g, is a smooth function

$$g:V(M) \times V(M) \rightarrow C^{\infty}(M)$$; $$\left\langle \textbf{X},\textbf{Y}\right\rangle \mapsto \left\langle\textbf{X},\textbf{Y}\right\rangle$$

satisfying the properties that for all $$\textbf{X}$$, $$\textbf{Y}$$ and $$\textbf{Z}\in V(M)$$, and for all $$f \in C^{\infty}$$:

1- $$\left\langle\textbf{X},\textbf{Y}\right\rangle= \left\langle\textbf{Y},\textbf{X}\right\rangle$$ (symmetry);
2a- $$\left\langle f\textbf{X},\textbf{Y}\right\rangle= f\left\langle\textbf{X},\textbf{Y}\right\rangle$$,
2b- $$\left\langle\textbf{X}+\textbf{Y},\textbf{Y}\right\rangle= \left\langle\textbf{X},\textbf{Z}\right\rangle + \left\langle\textbf{Y},\textbf{Z}\right\rangle$$ (bilinearity);
3- $$\left\langle\textbf{X},\textbf{X}\right\rangle>0$$ (positive definiteness);

Hope this helps.

AB

Last edited: Jan 24, 2010
3. Jan 24, 2010

### latentcorpse

hmmm. this is the spacetime metric. $\left\langle X, X \right\rangle$ isn't necessarily positive. what if X is a timelike vector? also, surely it should give me a real answer not an answer in $C^{\infty}(M)$
and if its the spacetime metric should i mention somewhere that its' covariant derivative vanishes?
thanks.

4. Jan 25, 2010

### Altabeh

In most standard textbooks where the geometric applications of tensors are discussed, the third property is included in the definition of metric tensor which is of course replaced by a weaker assumption $$|g_{ij}|\neq 0$$ where |..| is the determinant. By this property, you would be able to define ds2>0 or even ds2<0 because the arc length formula appears to be of the form

$$L=\int_{a}^{b}\sqrt{\left|g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}\right|}$$.

So the invariant line-element ds2 would be worked into the differential equation of metric as

$$\pm ds^2=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}$$.

So calling for $$g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}$$ to be larger than zero does not mean we are neglecting spacelike vectors dx^i/dt, does it?

A moral: if you are uncomfortable with $$C^{\infty}(M)$$, then use a weaker condition like $$C^{2}(M)$$ which is acceptable from the viewpoint of GR since in the Christoffel symbols, the metric tensor just appears to at most of the the order 2 in the derivatives wrt coordinates.

AB

Last edited: Jan 25, 2010
5. Jan 25, 2010

### latentcorpse

i'm also confused as to whether i sohuld be referring to vectors as $X$ or $X^a$. most textbooks appear to use the latter - should i do this also? it's just i though $X^a$ were the components of $X$, not the vector itself? to be honest it would probably be easier to refer to them as $X^a$ as that way i won't ahve to change it throughout my project. i guess i'm just looking for some justification on this.

secondly in various places i read $X^a \in M$ and then also $X^a \in V_p$. when we're working with vectors in GR, are they in the tangent space or the manifold?

i'm not uncomfortable with the notion of $C^{\infty}(M)$. i just don't understand why this is the codomain of $g$ and not $\mathbb{R}$?

thanks.

6. Jan 25, 2010

### Altabeh

1- Actually $X^a$ with a=0,1,2,3 in a 4D spacetime are the components of the vector $X$ and roughly speaking, they appear to be in the arrangement $$(X^0,X^1,X^2,X^3)=X$$. However, physicists call $X^a$ a contravariant vector but mathematicians don't. You keep the path of physicists, Weinberg as the most famous one among all, in what name you should assign to it and don't be worried about the outcome because everyone will definitely be acquainted with this new language when taking a look at your essay.

2- In the definition of metric, $X^i$ and $Y^j$ are chosen from the tangent space in the neighborhood of a given point, P, on the manifold M. So at an arbitrary point P in the manifold, any local vector $$X$$ lies in the tangent space $$V_p$$ or $$T_p$$ to the manifold at P. Indeed, $$V_p$$ consists of the set of all (local) vectors at the point P. To get a better visualization of the whole matter, we can go with the displacement vector $$X^{a}_Q-X^{a}_P$$ where $$X^a$$ represents the components of a vector $$X$$ in the tangent space at two points Q and P. The "displacement vector" connecting the points P and Q does not lie in the manifold M and therefore does not carry an intrinsic geometrical meaning. However, we can define the infinitesimal displacement vector $$dX^a$$ between two adjacent points P and Q, since this is a local quantity. When $$P\rightarrow Q$$, the vector $$dX^a$$ lies in the tangent space at P.

3- Since $g_{ab}X^aY^b$ is not always a number or something like that, rather it is just an invariant smooth (not necessarily up to infinity as in GR it could be up to the order 2) scalar function so it does belong to the class of functions $$C^{\infty}(M)$$. Though when normalized, it is always equal to 1 or -1.

Hope these are helpful to you!

Last edited: Jan 25, 2010