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Picture below is from do Carmo's Riemannian Geometry. I want to get the red part.

What I try:

Since $\frac{1}{n-1} Q(X,Y)$ is bilinear form, I can assume $$ \frac{1}{n-1} Q(X,Y) = \sum_{i,j} x^iy^j Q_{ij} $$ where $X=\sum_i x^iX_i, Y=\sum_j y^j X_j$. I think $Q_{ij}$ is the coefficients of $\frac{1}{n-1} Q$. Then, I have \begin{align} Q_{ij}&= \frac{1}{n-1} Q(X_i, X_j) \\ &= \frac{1}{n-1} \sum_k\langle R(X_i, X_k)X_j, X_k\rangle \\ &= \frac{1}{n-1} \sum_l R_{ikj}^l g_{lk} \text{ ~ ~ ~ or ~ ~ ~ } \frac{1}{n-1} \sum_k R_{ikjk} \end{align} Obviously, it is not same with the red part. How should I do it?

enter image description here

Enhao Lan
  • 5,829

1 Answers1

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I am currently self-studying this book and struggled quite a bit to understand that part. Here is a proof that I found:

Starting with some preliminaries, let $\{e_k\}$ denote an orthonormal basis of $T_p M$, then we can expand each element of $\{e_k\}$ in our original basis $\{X_i\}$:

\begin{equation} e_k = \sum_{i}{A_{ki}X_{i}} \end{equation}

By orthogonality of the $\{{e_k}\}$ we have:

\begin{equation} \delta_{ij} = \langle e_i, e_j \rangle = \langle \sum_{k}{A_{ik}X_k}, \sum_{l}{A_{jl}X_l} \rangle = \sum_{kl}{A_{ik} g_{kl} A_{jl}}, \end{equation}

Let $[A]$ be the matrix with $(i,k)$-th entry $A_{ik}$, and $[g]$ the matrix with entries $g_{ik}$. The above then gives:

\begin{equation} I = [A][g][A]^T \Longrightarrow [g]^{-1} = [A]^{T} [A] \end{equation}

If we let the $g^{ij}$ be the $(i,j)$-th entry of $[g]^{-1}$, we can write:

\begin{equation} g^{ij} = \sum_{k}{A_{ki} A_{kj}} = \sum_{k}{A_{kj} A_{ki}} = g^{ji} \end{equation}

Now we can show the expressions for the coefficient of the Ricci tensor in our local coordinate basis (the part highlighted in red in your post). Using the shorthand $R(a, b, c, d) = \langle R(a, b)c, d \rangle$:

\begin{equation} R_{ik} = Q(X_i, X_k) = \sum_{j}{R(X_i, e_j, X_k, e_j)} = \sum_{j}{R(X_i, \sum_{l}{A_{jl}X_{l}}, X_k, \sum_{m}{A_{jm}X_{m}})} \end{equation}

Taking out the summations and $A$ coefficients:

\begin{equation} = \sum_{jml}{A_{jl} A_{jm} R(X_i, X_{l}, X_k, X_{m})} = \sum_{jml}{A_{jl} A_{jm} R_{ilkm}} = \sum_{ml}{R_{ilkm} \sum_{j}{A_{jl} A_{jm}}} = \sum_{ml}{R_{ilkm} g^{ml}} \end{equation}

Re-labelling the indices in the last expression above gives us equality between the first and third expressions in the red box in your post. To show the equality between the second and third expressions, observe that:

\begin{equation} \sum_{sj}{R_{ijks} g^{sj}} = \sum_{sj}{\sum_{l}{R^{l}_{ijk} g_{ls} g^{sj}}} = \sum_{lj}{R^{l}_{ijk} \sum_{s}{g_{ls} g^{sj}}} = \sum_{lj}{R^{l}_{ijk} \delta_{lj}} = \sum_{j}{R^{j}_{ijk}} \end{equation}

Here, the justification for the first equality was shown on page 93, and the third was stated in the text above the red highlight.

rat123
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