I How Does the Metric of a 4-D Spacetime Define Its Symmetry and Expansion?

[Apashanka]

Our 4-D space is $x^1,x^2,x^3 ,t$.
Our sub-manifold is defined by $(x^1,x^2,x^3)$
Therefore for this sub-manifold to be maximally symmetric and for which the tangent vector $\frac{∂}{∂t}(\hat t)$ orthogonal to this sub-manifold
The metric becomes,
$ds^2=g(t)dt^2+f(t)(dr^2+r^2d\Omega^2)$
From the known metric for 4-D space and comparing this with above $g(t)=-1$ and for static $f(t)=1$ and for expanding $f(t)=a(t)^2$ ,is it the case ??
And can we say that the hypersurface to any t is orthogonal to the increament direction of t??

 

[PeterDonis]

the known metric for 4-D space

First, spacetime, not space. You got this right in the title of the thread.

Second, what metric for 4-D spacetime? There is no single "known" metric; it depends on what spacetime you want to look at.

comparing this with above $g(t)=-1$ and for static $f(t)=1$ and for expanding $f(t)=a(t)^2$ ,is it the case ??

Is what the case? What are you asking?

can we say that the hypersurface to any t is orthogonal to the increament direction of t??

If you mean, is every hypersurface of constant $t$ orthogonal to the vector $\partial / \partial t$, you should be able to answer that from the last thread we had on that subject.