Let $(M,g)=(N,\ddot{g})\times f(B,\bar{g})$ be an Einstein warped-product manifold Ricci flat (i.e. $Ric=\lambda g$ with $\lambda=0$) where $f:N \rightarrow (0, \infty)$ (positive scalar function) and with $g= \ddot{g}+f^2 \bar{g}$.


If $(B, \bar{g})$ is Ricci flat, being $(M, g)$ Ricci flat, this means that $(N, \ddot {g})$ must be only Ricci flat or not?



Einstein warped product manifold Ricci flat
Let $\bar M=M\_1\times\_f M\_2$ be a warped product manifold, $X\_1, Y\_1$ be two lifts of two vector fields on $M\_1$ to slides $M\_1\times p\_2$ and $X\_2, Y\_2$ be two lifts of two vector fields on $M\_2$ to slides $p\_1 \times M\_2$. Then  

$$\bar Ric(X\_1,Y\_1)= Ric^1(X\_1,Y\_1)-\frac{dimM\_2}{f}Hess^f(X\_1,Y\_1)$$  

$$\bar Ric(X\_2,Y\_2)= Ric^1(X\_2,Y\_2)-f^\*g\_2(X\_2,Y\_2)$$  

where $f^\*=f\Delta f-(dimM\_2-1)g\_1(grad f, grad f)$, $Ric^i$ is the lift of the Ricci curvatire tensor to $M\_i$ and $\bar Ric$ is the Ricci curvature of $\bar M$. Now assume that $\bar M$ and $M\_1$ are Ricci flat, then  

$$0= -\frac{dimM\_2}{f}Hess^f(X\_1,Y\_1)$$  

$$0= Ric^2(X\_2,Y\_2)-f^\*g\_2(X\_2,Y\_2)$$  

The first equation implies that $Hess^f$ is zero i.e. the gradient of $f$ is a constant say $c$ and consequently  

$$f^\*=f\Delta f-(dimM\_2-1)g\_1(grad f, grad f)=-(dimM\_2-1)c^2$$  

Thus the second factor is Einstein.
Now assume that $\bar M$ and $M\_2$ are Ricci flat, then  

$$0= Ric^1(X\_1,Y\_1)-\frac{dimM\_2}{f}Hess^f(X\_1,Y\_1)$$  

$$0= -f^\*g\_2(X\_2,Y\_2)$$  

Then the first equation yields by tracing 
$$r\_1=\frac{dimM\_2}{f}\Delta f$$ 
and consequently the second equation implies  

$$ f^\*=f\Delta f-(dimM\_2-1)g\_1(grad f, grad f)=0$$  

i.e. 
$$r\_1f^2=dimM\_2(dimM\_2-1)g\_1(grad f, grad f)$$


