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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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up vote 1 down vote accepted

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)

The full answer can be found, for instance, on p. 23 of the beautiful book of C.Hopper and B.Andrews "The Ricci flow in Riemannian geometry" (can be downloaded from B.Andrews site)

Another reference that may be equally useful (with more information provided) is in Proposition 13.4 on p.404 and subsequent remarks here.

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Thanks! The references and info you provided are really useful. I did exercise 25 in chapter of Petersen which was similar to proving the second identity used in proving Berger's Lemma (in your first reference) but didn't see the connection to this exercise. – Bohring Oct 18 '12 at 4:39
I let the info digest for a few days and I feel confident that I understand the proof of Berger's Lemma, and how to use polarization to show that the sectional curvature determines the Riemannian curvature, but I am perhaps failing to understand the curvature operator. I feel like this is a rather naive question and that I have a bit more work to do, but is the norm of the curvature operator the same as the norm of the curvature tensor? – Bohring Oct 20 '12 at 21:16
@Pete See Petersen's "Riemannian geometry", Chapter 2, section 6.3 "Norms of Tensors" where the author give the convention for the choice of norms. In brief, it is the Euclidean norm, that comes from the inner product. – Yuri Vyatkin Oct 21 '12 at 0:01
@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

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