# Bounding the Norm of the Riemann Curvature Operator

I am having trouble with exercise 26 in chapter 2 of Peter Petersen's text "Riemannian Geometry." The exercise is stated:

"Using Polarization show that the norm of the curvature operator on $\Lambda^2 T_pM$ is bounded by $$|\mathcal{R}|_p \leq c(n)|\text{sec}|_p$$ for some constant $c(n)$ depending on dimension and where $|\text{sec}|_p$ denotes the largest absolute value for any sectional curvature of a plane in $T_pM$."

I understand how to write the norm of the curvature tensor as $|R|^2 = R_{ijk}^l R^{ijk}_l$. And I see that \begin{eqnarray*} R_{ljj}^l &=& g(R(\frac{\partial}{\partial x^l},\frac{\partial}{\partial x^j})\frac{\partial}{\partial x^j},\frac{\partial}{\partial x^l})\\ &=& \sum^{n}_1 c \sec(\frac{\partial}{\partial x^j},\frac{\partial}{\partial x^l})\\ &\leq& c(n)|\text{sec}|_p, \end{eqnarray*} where $c$ is some constant, but I'm not sure how to proceed or in what way the author intends us to use polarization...

EDIT (Attempt at a Solution): Since $\mathcal{R}$ is self adjoint, there exists an orthonormal basis for $\Lambda^2 T_pM$ consisting of eigenvectors of $\mathcal{R}$. Let $v_1,...,v_n$ be such a basis, i.e. $\mathcal{R}v_i = \lambda_i v_i, \; g(v_i,v_j) = 0 \; \text{and} \; ||v_i|| = 1$. We prove this using the operator norm defined in the text $|\mathcal{R}|_p = \max\{|\lambda_j|\}$ where $\lambda_j$ is an eigenvalue of $\mathcal{R}.$ Let $|\lambda_k| = \max\{|\lambda_j|\}$ \begin{eqnarray*} |\mathcal{R}|^2_p &=& (\max\{|\lambda_j|\})^2\\ &=& |\lambda_{k}|^2\\ &=& |g(\mathcal{R}(v_k),\mathcal{R}(v_k)|\\ &=& |g(\mathcal{R}(v_k),\lambda_k v_k|\\ &=& |\lambda_{k}||g(\mathcal{R}(v_k), v_k)|\\ &=& |\lambda_{k}| |\sec{(v_k)}|\\ &\leq& |\lambda_{k}| |\sec|_p.\\ \end{eqnarray*} Thus, $|\mathcal{R}|_p = |\lambda_{k}| \leq |\sec|_p$. Then, by equivalence of norms we obtain $|\mathcal{R}|_p| \leq c(n)|\sec|_p$ for the Euclidean norm, where $c(n)$ is some constant $c(n)$ depending on dimension.

The problem with the above is that the $v_i$ might not be simple ($v_i$ might not equal $x_1 \wedge x_2$ where $x_1,x_2 \in T_PM$) so that $g(\mathcal{R}(v_k), v_k)$ isn't a sectional curvature....

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By "polarization" the author understands here the process of computing the sectional curvature $K(X \wedge Y)$ on the plane $$\tfrac{1}{2}(e_i + e_k)\wedge(e_j + e_l)$$ that gives an expression of the Riemannian $R_{i j k l}$ in terms of $K(X \wedge Y)$ only! (As it is well known the sectional curvature completely determines the Riemannian curvature)
@Pete Also, I had a chance to open the book, and to look at the exercises. The previous one (25) gives another hint how to proceed. Also, I noticed that the polarization is explained in the section on Sectional curvature as $w = w_1 + w_2$. – Yuri Vyatkin Oct 21 '12 at 0:06
Thanks again for your help. I saw the section on Norms and was able to calculate $|R|^2$. I'm going to give it another try tonight and see if I can figure this out or at least pinpoint exactly what I am missing. – Bohring Oct 21 '12 at 4:54