# Scalar Curvature and Connected Sum

Given that I know the scalar curvatures of two compact manifolds, is there anything I can say about the scalar curvature of the connected sum? For my specific application my compact manifolds are actually Ricci-Flat so would it be possible to say that the scalar curvature of the connected sum vanishes?

First of all, given connected Riemannian manifolds $$(M_1, g_1)$$ and $$(M_2, g_2)$$, the connected sum $$M_1\# M_2$$ does not admit a canonical metric, even if you require it to 'extend' $$g_1$$ and $$g_2$$. So instead, one should ask if $$M_1\# M_2$$ admits a metric with certain properties.
Let $$M_1$$ and $$M_2$$ be connected manifolds which admit positive scalar curvature metrics. If $$\dim M_1 = \dim M_2 \geq 3$$, then $$M_1\# M_2$$ admits metrics of positive scalar curvature.
As for your specific application, it is not necessarily the case that the connected sum of Ricci flat manifolds admits a metric of zero scalar curvature. For example, a $$K3$$ surface has a Ricci flat metric, but $$K3\# K3$$ does not admit a metric of zero scalar curvature.
To see this, note that $$K3\# K3$$ is a spin manifold with
$$\widehat{A}(K3\# K3) = -\frac{1}{8}\sigma(K3\# K3) = -\frac{1}{8}(-32) = 4 \neq 0,$$ so by Lichnerowicz's Theorem, $$K3\# K3$$ does not admit a metric of positive scalar curvature. On a manifold without positive scalar curvature metrics, any non-negative scalar curvature metric is actually Ricci-flat, so if $$K3\# K3$$ were to admit a scalar flat metric, it must be Ricci flat. In particular, $$K3\# K3$$ would have an Einstein metric, but this is impossible as $$\chi(K3\# K3) = 46$$ and $$\sigma(K3\# K3) = - 32$$ violate the Hitchin-Thorpe inequality $$\chi(M) \geq \frac{3}{2}|\sigma(M)|$$.