[proofplan] We write $R:=\operatorname{Rm}$ and estimate the components of $Q(R)$ in a fixed [orthonormal basis](/page/Orthonormal%20Basis). Each component is a sum of three contracted quadratic expressions, and each contracted sum is bounded by $|R|^2$ using the finite-dimensional [Cauchy-Schwarz inequality](/theorems/432) on the index set $\{1,\dots,n\}^2$. This argument does not use the algebraic curvature symmetries, so it proves the same quadratic estimate for any $4$-tensor for which the displayed Hamilton reaction formula is defined. After obtaining a uniform componentwise bound, we sum over all four free indices to get the tensor norm estimate with a positive constant depending only on $n$. [/proofplan]

[step:Fix an orthonormal basis and write the induced tensor norms] At the point at which the estimate is being evaluated, let $V$ denote the underlying $n$-dimensional real tangent space and let $g: V \times V \to \mathbb{R}$ denote the Riemannian [inner product](/page/Inner%20Product) on $V$. Write $R:=\operatorname{Rm}$ for the curvature tensor appearing in the theorem statement, regarded as a multilinear map $R: V \times V \times V \times V \to \mathbb{R}$. Let $(e_1,\dots,e_n)$ be a $g$-orthonormal basis of $V$. For indices $i,j,k,l,p,q \in \{1,\dots,n\}$, define $R_{ijkl} := R(e_i,e_j,e_k,e_l)$ and $Q(R)_{ijkl} := Q(R)(e_i,e_j,e_k,e_l)$. Because the basis is orthonormal, the induced tensor norms are \begin{align*} |R|^2 = \sum_{a,b,c,d=1}^n R_{abcd}^2. \end{align*} Also, \begin{align*} |Q(R)|^2 = \sum_{i,j,k,l=1}^n Q(R)_{ijkl}^2. \end{align*} [/step]

[step:Bound each contracted quadratic sum by $|R|^2$]Fix indices $i,j,k,l \in \{1,\dots,n\}$. Define three [real numbers](/page/Real%20Numbers) by \begin{align*} A_{ijkl} := \sum_{p,q=1}^n R_{ipkq}R_{jplq}. \end{align*} Define \begin{align*} B_{ijkl} := \sum_{p,q=1}^n R_{iplq}R_{jpkq}. \end{align*} Define \begin{align*} D_{ijkl} := \sum_{p,q=1}^n R_{ijpq}R_{klpq}. \end{align*} Applying the Cauchy-Schwarz Inequality to the Euclidean inner product on $\mathbb{R}^{n^2}$ gives \begin{align*} |A_{ijkl}| &\le \left(\sum_{p,q=1}^n R_{ipkq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{jplq}^2\right)^{1/2} \le |R|^2. \end{align*} For $B_{ijkl}$, applying the Cauchy-Schwarz Inequality to the arrays $(R_{iplq})_{p,q}$ and $(R_{jpkq})_{p,q}$ in $\mathbb{R}^{n^2}$ gives \begin{align*} |B_{ijkl}| &\le \left(\sum_{p,q=1}^n R_{iplq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{jpkq}^2\right)^{1/2} \le |R|^2. \end{align*} For $D_{ijkl}$, applying the Cauchy-Schwarz Inequality to the arrays $(R_{ijpq})_{p,q}$ and $(R_{klpq})_{p,q}$ in $\mathbb{R}^{n^2}$ gives \begin{align*} |D_{ijkl}| &\le \left(\sum_{p,q=1}^n R_{ijpq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{klpq}^2\right)^{1/2} \le |R|^2. \end{align*}[/step]

[guided]Fix $i,j,k,l \in \{1,\dots,n\}$. The component formula for $Q(R)_{ijkl}$ contains three sums over the contracted indices $p$ and $q$, so we isolate them. Define \begin{align*} A_{ijkl} := \sum_{p,q=1}^n R_{ipkq}R_{jplq}. \end{align*} Define \begin{align*} B_{ijkl} := \sum_{p,q=1}^n R_{iplq}R_{jpkq}. \end{align*} Define \begin{align*} D_{ijkl} := \sum_{p,q=1}^n R_{ijpq}R_{klpq}. \end{align*} The point is that each of these is an ordinary dot product of two vectors indexed by the finite set $\{1,\dots,n\}^2$. For example, define two vectors $u,v \in \mathbb{R}^{n^2}$ by $u_{pq} := R_{ipkq}$ and $v_{pq} := R_{jplq}$. Then $A_{ijkl}=\sum_{p,q=1}^n u_{pq}v_{pq}$. By the Cauchy-Schwarz Inequality applied in the finite-dimensional Euclidean space $\mathbb{R}^{n^2}$, \begin{align*} |A_{ijkl}| \le \left(\sum_{p,q=1}^n u_{pq}^2\right)^{1/2} \left(\sum_{p,q=1}^n v_{pq}^2\right)^{1/2}. \end{align*} Substituting $u_{pq}=R_{ipkq}$ and $v_{pq}=R_{jplq}$ gives \begin{align*} |A_{ijkl}| \le \left(\sum_{p,q=1}^n R_{ipkq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{jplq}^2\right)^{1/2}. \end{align*} Each displayed sum is a partial sum of the full squared tensor norm \begin{align*} |R|^2=\sum_{a,b,c,d=1}^n R_{abcd}^2. \end{align*} Therefore both factors are at most $|R|$, and hence \begin{align*} |A_{ijkl}| \le |R|^2. \end{align*} Now estimate $B_{ijkl}$ without suppressing the indices. Define two vectors $s,t \in \mathbb{R}^{n^2}$ by $s_{pq}:=R_{iplq}$ and $t_{pq}:=R_{jpkq}$. Then $B_{ijkl}=\sum_{p,q=1}^n s_{pq}t_{pq}$, so the Cauchy-Schwarz Inequality in $\mathbb{R}^{n^2}$ gives \begin{align*} |B_{ijkl}| &\le \left(\sum_{p,q=1}^n R_{iplq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{jpkq}^2\right)^{1/2} \le |R|^2, \end{align*} because both sums are partial sums of $\sum_{a,b,c,d=1}^n R_{abcd}^2$. Finally estimate $D_{ijkl}$. Define two vectors $m,r \in \mathbb{R}^{n^2}$ by $m_{pq}:=R_{ijpq}$ and $r_{pq}:=R_{klpq}$. Then $D_{ijkl}=\sum_{p,q=1}^n m_{pq}r_{pq}$, and Cauchy-Schwarz gives \begin{align*} |D_{ijkl}| &\le \left(\sum_{p,q=1}^n R_{ijpq}^2\right)^{1/2} \left(\sum_{p,q=1}^n R_{klpq}^2\right)^{1/2} \le |R|^2, \end{align*} again because both factors are bounded above by $|R|$.[/guided]

[step:Estimate every component of $Q(R)$] We use the Hamilton curvature reaction convention in the fixed $g$-orthonormal basis: for every $i,j,k,l \in \{1,\dots,n\}$, \begin{align*} Q(R)_{ijkl} &:= 2\sum_{p,q=1}^n R_{ipkq}R_{jplq} -2\sum_{p,q=1}^n R_{iplq}R_{jpkq} +\sum_{p,q=1}^n R_{ijpq}R_{klpq}. \end{align*} With the definitions of $A_{ijkl}$, $B_{ijkl}$, and $D_{ijkl}$ from the previous step, this is exactly \begin{align*} Q(R)_{ijkl}=2A_{ijkl}-2B_{ijkl}+D_{ijkl}. \end{align*} Using the triangle inequality and the bounds from the previous step, \begin{align*} |Q(R)_{ijkl}| \le 2|A_{ijkl}|+2|B_{ijkl}|+|D_{ijkl}| \le 2|R|^2+2|R|^2+|R|^2 = 5|R|^2. \end{align*} [/step]

[step:Sum the componentwise estimate to obtain the tensor norm bound] Squaring the componentwise bound and summing over all free indices gives \begin{align*} |Q(R)|^2 = \sum_{i,j,k,l=1}^n Q(R)_{ijkl}^2 \le \sum_{i,j,k,l=1}^n 25 |R|^4 = 25 n^4 |R|^4. \end{align*} Taking square roots yields \begin{align*} |Q(R)| \le 5n^2 |R|^2. \end{align*} Thus the theorem holds with the explicit dimension-dependent constant \begin{align*} C_n := 5n^2. \end{align*} [/step]