u r not right dear Low determinant is n- li,near form n-tensor and it is a scalar...

@Cenap Özel. The invariant way to think of the determinant det(g) of the metric tensor is as section of

(\wedge^{max} T^*M)^2,

and is certainly not a scalar.  @Robert Low is perfectly right in saying that its coordinate expression depends on the coordinate system and, to put it bluntly, being constant does not really make sense (@Robert is just too polite to put it so bluntly). Considered in this way it is  always a parallel section with respect to the Levi-Civita connection, which is the closest thing to being constant that does make sense.  This is used all the time when we take the positive square root \sqrt(det(g)) which is a nowhere vanishing section of

|\wedge^{max}| T^*M := (\wedge^max T^*M)\tensor flat-orientation-bundle 

i.e. the real line bundle of densities. Here the flat-orientation bundle is the real line bundle with the constant transition functions sign(det (transition-matrix)). It defines the Riemannian measure aka its volume element. That is where the ubiquitous \sqrt(det(g))  comes from in physics (usually with an extra minus because of the use of Lorentzian geometry).

@Robert You write that  det(g) defines the volume form, but that is actually  sqrt(det(g)) as I am sure you know, and as I explain above.

Yes you can find such a metric but in a certain local coordinate system as @Robert  and @Rogier answered in many details. But if you give us many details about the request as why are you searching for such metric? this me helps out.

We are working on gauge theories..

Sameh shenawy: if we take two times covariant derivative of Levi-Civata connection it gives us Laplacian with determjnant of metric. So if Laplacian acts on Ricci scalar bc of Leibniz rule this determinant gives some gradient terms. So to eliminate these terms determinat must be constant. Another question: What is eigenvalue of the Laplacian operator on Ricci scalar?

Dear Cenap

Sure for example (R^n,id) and in R^3 for example all planes and also cylinder (in cyl. coords) have metric tensor G=diag(1,1) so that det(G)=1 

I also refer to previous answer: Sure in a different chart the matrix G of riemannian metric is most likely different, but I am guessing the question involved metric G in a fixed chart ?

Was this your question perhaps I did not understand it correctly ? ?

regards

Samuli

 thanks dear Samuliiii