What is the point of sub and superscripts in metric tensor I am new to QFT and have to catch up with some things that I did not learn from my SR courses. I am learning about the tensor notation, in particular the metric tensor, $$g_{\nu\mu}=g_{\mu\nu}=\begin{bmatrix} 1 & 0 & 0 & 0 \\0 & -1 & 0 & 0 \\0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$
However, I don't understand the point of the subscripts, as the content of the matrix seems to be independent of both $\mu$ and $\nu$. So what do those sub indexes indicate? 
 A: Writing
$$M_{\mu \nu} = \pmatrix{a & b \\ c & d}$$
is shorthand notation for
$$M_{00} = a, M_{01} = b, M_{10} = c, M_{11} = d$$
In this sense, you're right - the subscripts $\mu\nu$ don't really have any particular significance in this expression.  It would be not be unreasonable to simply write
$$\mathbf M = \pmatrix{a & b \\ c & d}$$

However, it is of utmost importance to remember that there is a difference between a tensor $\mathbf M$ - a basis-independent geometrical object - and the components of that tensor $M_{\mu \nu}$ (where $\mu,\nu = 0,1,\ldots $) as expressed in a particular basis.  I personally tend to attach the indices to make it extra clear that I'm working in some basis, while any object in bold font with no indices should be understood to be basis-independent, but that's just a personal convention.
A: In SR, the invariant "length" of a vector is given by 
$$
ds^2= (c dt)^2-(dx^2+dy^2+dz^2)
$$
and so you need to keep track of where the signs appear.  Thus, it is convenient to introduce a metric $g_{\mu\nu}$ so that, if $(cd t,dx,dy,dz)$ are components of a vector $(dx^0,dx^1,dx^2,dx^3)$ then 
$$
ds^2= dx^\mu g_{\mu\nu} dx^\nu
$$
with $g_{\mu\nu}$ as given.  Since expressions such as $g_{\mu\nu} dx^\nu$ occur quite often, one defines $dx_\mu=g_{\mu\nu} dx^\nu$ so that we can also write
$$
ds^2=dx^\mu dx_\mu\, ,
$$
where $dx_0=c dt, dx_1=-dx, dx_2=-dy, dx_3=-dz$.  
The raising and lowering of indices is done with $g_{\mu\nu}$ or its inverse $g^{\mu\nu}$ (they happen to coincide in your case but might not when using non-Cartesian bases for 3-space).
Note that in ordinary Euclidean space, $\tilde g_{ij}=\hbox{diag}(1,1,1)$ and the length is just
$$d\ell^2= dx^2+dy^2+dz^2 = dx^i \tilde g_{ij}dx^j.$$  Since $\tilde g$ is just the unit matrix, there is 
no difference between $dx^i$ and $dx_i$.  
In general you need a metric when the scalar product between two vectors is "non-standard".  This happens in SR because of the - signs in the $ds$ but you also need a metric if you are using non-orthogonal coordinate system.  In such system (say in the 2-d place where $\hat x$ and $\hat y$ are not orthogonal) one can write a general vector
$$
\vec V=V^x\hat x+V^y\hat y
$$
meaning you take a specific linear combo of $\hat x$ and $\hat y$ to get $\vec V$.  However,  $V^x$ and $V^y$ are not components obtained by the scalar product of $\vec V$ with $\hat x$ or $\hat y$.  Rather
$$
V_x=\vec V\cdot\hat x\, ,\qquad V_y=\vec V\cdot\hat y
$$ 
and $V_x=V^x+V^y\cos\alpha$ where $\alpha$ is the angle between $\hat x$ and $\hat y$.  You can find more details on this nice geometrical distinction between components with upper and lower indices in 

James Evans, Am. J. Phys. 65  (1997) p. 1039.

I'm including a diagram from this paper as a teaser, showing the meaning of $V^x, V^y, V_x$ and $V_y$:

In GR, metric tensors are often non-diagonal.
Finally, there is a nice discussion done in the context of SR in the (wonderful) book 

Brau, C.A., 2004. Modern problems in classical electrodynamics. Oxford univ. press

with various examples, including the electromagnetic field tensor $F^{\mu\nu}$.  The positioning of indices in this tensor is particularly important.
A: The indices indicate the value of the matrix. For example g00 = -1, g01=0. The difference between sub and superscripts is the two different types tensors. Covariant and contravariant. You can figure out the other one by the equation gνμ(covariant)*gνμ(contravariant)=(identity-matrix)νμ
Also note that g00 should be c^2. The identity matrix is just the metric tensor except g00-> gnn is 1
