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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?

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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.

One example where you can say something, is a theorem of Gromov and Lawson:

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 $\chi(M) \geq \frac{3}{2}|\sigma(M)|$.

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