📂Geometry

Scalar Curvature of Differential Manifolds

Definition¹

Riesz Representation Theorem

Let $\left( H, \left\langle \cdot,\cdot \right\rangle \right)$ be an inner product space. For linear functionals $f \in H^{ \ast }$ and $\mathbf{x} \in H$ on $H$, there exists a unique $\mathbf{w} \in H$ that satisfies $f ( \mathbf{x} ) = \left\langle \mathbf{w} , \mathbf{x} \right\rangle$.

Let $M$ be a differentiable manifold. Let $T_{p}M$ be the tangent space at point $p\in M$. Now, for a fixed $X \in T_{p}M$, consider a linear functional $\Ric(X, \cdot)$ on $T_{p}M$. If $\Ric$ is the Ricci curvature, then by the Riesz Representation Theorem, there exists a unique $Z$ that satisfies the following in terms of $Y$ and $\Ric$.

$$\Ric (X, Y) = g\left( Z, Y \right)$$

Now, let us define $K : T_{p} \to T_{p}M$ for such $X, Z$ as follows.

$$K(X) = Z$$

Then

$$\Ric (X, Y) = g\left( K(X), Y \right)$$

and the scalar curvature $K : M \to \mathbb{R} \text{ by } K(p) = K_{p}$ at point $p$ is defined as follows.

$$K_{p} = \text{Trace of } K$$

Explanation

Let $\left\{ X_{i} = \dfrac{\partial }{\partial x_{i}}\right\}$ be the basis of $T_{p}M$. Then, by the trace inner product formula,

$$K_{p} = \tr(K) = g(K(X_{i}), X_{j})g^{ij} = \Ric (X_{i}, X_{j}) g^{ij} = R_{ikj}^{k}g^{ij}$$

Since $R_{ikj}^{k} = R_{ikj}^{s}\delta_{s}^{k} = R_{ikj}^{s}g_{sl}g^{lk} = R_{ikjl}g^{lk}$ is so,

$$K_{p} = R_{ikj}^{k}g^{ij} = R_{ikjl}g^{lk}g^{ij}$$

Therefore, just as the Ricci curvature is the average of the second and fourth components of the Riemann curvature, the scalar curvature is the average over all components of the Riemann curvature. Especially, if $\left\{ Z_{i} \right\}$ are the orthonormal basis of $T_{p}M$, then since $g^{ij} = \delta_{ij}$,

$$K_{p} = \Ric(Z_{i}, Z_{j})\delta_{ij} = \Ric(Z_{i}, Z_{i}) = R(Z_{i}, Z_{j}, Z_{i}, Z_{j})$$

holds, and this is equivalent to the average of the sectional curvature $K(Z_{i}, Z_{j})$.