Why the metric is covariantly constant?

[Giammy85]

you can show that the metric is covariantly constant by writing:
V_a;b=g_acV^c;b

for linearity V_a;b=(g_acV^c);b=g_ac;bV^c+g_acV^c;b

than must be g_ac;b=0

is there an alternative argument (even shorter than this) that show that the metric is covariantly constant?

if I calculate g_ac;b considering that g_ac is a (0,2) tensor than I will write the 2 connections in form of the metric, but I have obtained this form using the fact that g_ac;b=0 so it seems to me like I'm just turning around

any help?

 

[shoehorn]

you can show that the metric is covariantly constant by writing:
V_a;b=g_acV^c;b

for linearity V_a;b=(g_acV^c);b=g_ac;bV^c+g_acV^c;b

than must be g_ac;b=0

This isn't a proof that the metric is covariantly constant because in the first step you have assumed that $g_{ac;b}=0$, which is equivalent to assuming that the metric is covariantly constant.

More formally, saying that the metric is covariantly constant means that the inner product between vectors is constant along a path of parallel transport. In particular, suppose that $X,Y$ are two vector fields, $g$ is a metric, and $\nabla$ is a connection. Then suppose that $V$ is a tangent vector to a path along which $X$ and $Y$ are parallel transported. If the metric is covariantly constant we must have

$$\nabla_Vg(X,Y) = V^iX^jY^k(\nabla_ig)_{jk} = 0$$

Since this holds for any such vectors and any path of parallel transport, the condition for the metric to be covariantly constant is

$$\nabla_ig_{jk}=0$$

 

[robphy]

Strictly speaking, I think that you have to spell out the meaning of ; , i.e., specify the derivative operator (or the connection). At face value, $$\nabla_a g_{bc}$$ is a nonzero tensor of type (0,3) that could be used to define a "nonmetricity tensor" [which is zero in the case of semi-Riemannian geometry].

 

[Giammy85]

;b is the covariant derivative with respect to b

 

[robphy]

;b is the covariant derivative with respect to b

That is not a sufficient definition.
What are its DEFINING properties (as you are presenting them)?
Without them, you can't prove anything.

What is the level of this course you are taking?
Maybe the answer to your question (with what you have given) is the Leibniz rule.

 

[Giammy85]

1 What are its DEFINING properties (as you are presenting them)?

2 What is the level of this course you are taking?

3 Maybe the answer to your question (with what you have given) is the Leibniz rule.

1 in general for a tensor T of upper indices a... and lower indices b...
T^a..._b...;c=T^a..._b...,c+GAMMA^a_dc*T^d..._b...+...-GAMMA^d_bc*T^a..._d...-...

So for (0,1) tensor: V_a;b=V_a,b-GAMMA^c_ab*V_c...

where GAMMA is the connection and ,b indicates the derivative with respect to b.

2 It's a Master course for Physics and Mathematics

3 I've used the Leibniz rule i.e property of linearity in what I have written in the first post. Now I need an alternative and shorter argument

 

[shoehorn]

1 in general for a tensor $T^{a_1\ldots a_r}_{\phantom{a_1\ldots a_r}b_1\ldots b_s}$

$$T^{a_1\ldots a_r}_{\phantom{a_1\ldots a_r}b_1\ldots b_s;c} = T^{a_1\ldots a_r}_{\phantom{a_1\ldots a_r}b_1\ldots b_s,c} + \sum_{\alpha=1}^r \Gamma^{a_\alpha}_{\phantom{a_\alpha}cd} T^{a_1\ldots a_{\alpha-1}da_{\alpha+1}\ldots a_r}_{\phantom{a_1\ldots a_{\alpha-1}da_{\alpha+1}\ldots a_r}b_1\ldots b_s} - \sum_{\beta=1}^s \Gamma^d_{\phantom{d}cb_\beta} T^{a_1\ldots a_r}_{\phantom{a_1\ldots a_r}b_1\ldots b_{\beta-1}db_{\beta+1}\ldots b_s}$$

So for a $(0,1)$ tensor,

$$V_{a;b} = V_{a,b} - \Gamma^c_{\phantom{c}ab}V_c$$

where $\Gamma^a_{\phantom{a}bc}$ are the connection coefficients and $_{,b}$ indicates the coordinate partial derivative with respect to $x^b$.

I've taken the liberty of translating the above into proper $\TeX$ (and more importantly, what I've written is correct: what you had written was not).

You used this expression for the covariant derivative to "prove" that the metric was covariantly constant. I've pointed out to you that this isn't a proof at all since in your "proof" you made the implicit assumption that $g_{ab;c}=0$ -- which is precisely the thing that you were trying to prove!

3 I've used the Leibniz rule i.e property of linearity in what I have written in the first post. Now I need an alternative and shorter argument

I've already explained to you that the argument you gave in the first post is incorrect. What's more, I've actually given you a correct argument in my first reply. In case this isn't enough for you, see any book on elementary differential geometry such as Nakahara or Frankel.

Alternatively, you could have used the search function to find posts in which the defining properties of the connection are discussed, such as this post in this thread.

 

[Giammy85]

Here the solution of my teacher:

Since g_ac is a (0,2) tensor than g_ac;b= g_ac,b-GAMMA^d_ab*g_dc-GAMMA^d_cb*g_ad

In a Local Inertial Frame g_ac,b=0 in a point P, than since the connection depends on the first partial derivatives of the metric GAMMA^d_ab=GAMMA^d_cb=0 in P.

So we have in a LIF: g_ac;b=0 in P
This is a tensor equation and it is valid in all frames since we have demonstrated its validity in one frame (LIF).

So g_ac;b=0 in all frames