Estimating scalar curvature by norm of Riemannian curvature tensor under the Ricci flow

Question

In B. Chow and D. Knopf's book "The Ricci Flow: An Introduction", the authors claim that for any dimension $n$ and any Riemannian manifold $M^n$, there is a constant $C_n$ depending only on $n$ such that $R \leq C_n \|\text{Rm}\|$. It's not clear whether this inequality should actually be $R(t) \leq C_n \|\text{Rm}(t)\|$, where $g(t)$ is a solution to the Ricci flow on $M$. It appears that the definition of norm they're using is the pointwise one, i.e $\|\text{Rm}(x)\| = \sqrt{R_{ijk\ell}(x)R^{ijk\ell}(x)}$.

Why is this true? From the definitions it doesn't look at all obvious why there should exist such a constant.

Answer 1

Remark

As @OthisChodosh and @WillieWong have pointed out, the existence of a constant $C_n$ that depends only on the dimension can be proved using only elementary linear algebra. I might as well provide the details. Although I like my first answer, it was overkill.

Simpler answer

First, recall that if $V$ is a finite dimensional inner product space, then given any linear function $\ell: V \rightarrow \mathbb{R}$, there exists a constant $C_\ell > 0$ such that, for any $v \in V$, $$ |\ell(v)| \le C_\ell |v|. $$ Given the tangent space $T$ at a point in a Riemannian manifold, let $$ V = S^2(\bigwedge^2V^*). $$ The inner product on $T$ induces an inner product on $V$, namely the one used here. The definition of scalar curvature defines a linear function $S: V\rightarrow \mathbb{R}$, and therefore there is a constant $C$ such that $$ |S(v)| \le C|v|. $$ Now note that the definition of $S$ is canonical, defined using only the dimension and inner product on $T$. The constant has to be independent of the inner product, because, if you write everything out above with respect to an orthonormal basis, the formulas all remain the same, no matter what inner product is, and therefore the constant $C$ does, too. Therefore, the constant $C$ depends on the dimension only.

Calculation of sharp constant You can compute the best possible value of $C_n$.

First, given two curvature-like tensors $A$ and $B$, let $$ A\cdot B = g^{is}g^{jt}g^{ku}g^{lv}A_{ijkl}B_{stuv}. $$ Then $|A|^2 = A\cdot A$.

The idea is to decompose the curvature tensor into two terms, $$ R_{ijkl} = T_{ijkl} + U_{ijkl}, $$ where $$ T_{ijkl} = \frac{S}{n(n-1)}(g_{ik}g_{jl}-g_{il}g_{jk}) $$ and $S$ is the scalar curvature. A straightforward calculation shows that $$ T\cdot R = T\cdot T = \frac{2S^2}{n(n-1)}, $$ which implies that $$ T\cdot U = T\cdot(R-T) = 0. $$ This, in turn, implies that $$ |R|^2 = |T|^2 + |U|^2 \ge |T|^2 = \frac{2S^2}{n(n-1)} $$ Therefore, $$ |S| \le \sqrt{\frac{n(n-1)}{2}}|R|, $$ with equality holding if and only if $$ R_{ijkl} = \frac{S}{n(n-1)}(g_{ik}g_{jl}-g_{il}g_{jk}). $$ If this holds everywhere on $M$, then the metric has constant sectional curvature.