Ricci Curvature Does Not Depend on the choice of orthonormal basis

Question

Let $x=z_{n}$ be a unit vector in $T_{p}M$ and take an orthonormal basis $\{z_1,z_2,...z_{n-1}\}$ of the hyperplane in $T_pM$ orthogonal to $x$.

Ricci curvature is defined as

$Ric_{p}(x)=\frac{1}{n-1}\Sigma_{i}\langle R(x,z_{i})x,z_{i} \rangle$

do Carmo proves that this is independent of the choice of orthonormal basis by defining a bilinear form $Q$ on $T_pM$ where for $x,y \in T_p M$

$Q(x,y)=$ trace of the mapping $z \mapsto R(x,z)y$.

Here is where I get lost:

Choosing $x$ a unit vector and then completing it to an orthonormal basis $\{z_1,z_2,...z_{n}=x\}$ of $T_p M$ we have

$Q(x,y)=\Sigma_{i}\langle R(x,z_i)y,z_i \rangle=\Sigma_i \langle R(y,z_i)x,z_i \rangle = Q(y,x)$. ***

So Q is symmetric and $Q(x,x)=(n-1)Ric_p(x)$ is intrinsically defined.

I am particularly confused about the line that I put *** by and what he means by "intrinsically defined".

Answer 1

What you want to verify is that the Ricci curvature is invariant under Riemannian isometries. $Q$ is the Ricci tensor. It is intrinsically defined in the sense that once the (Levi-Civita) connection is fixed, the Ricci tensor can be computed from it using only the metric (for example, by the method used to compute Q above, which required only the metric). Note that by the fundamental theorem of Riemannian geometry there is a well defined correspondence between Levi-civita connections and Riemannian metrics (uniquely computable). So the connection is invariant under isometry. (If $F:(M,g_M)\to (N,g_N)$ is a Riemannian isometry, then $g_M=F^{\ast}g_N$).

In other words, if you imbed (via isometry) your Riemannian manifold into another, the object you defined above (now in the ambient space) will assume the same value when evaluated on corresponding tangent vectors to corresponding points of the imbedding.

Answer 2

He is using the symmetry by pairs $R(X,Y,Z,W) = R(Z,W,X,Y)$ of the Riemann curvature. This tells you that $\langle R(x,z_i)y,z_i\rangle = \langle R(y,z_i)x,z_i\rangle$ for all $i=1,\ldots,n-1$. Sum that over $i$ and get $Q(x,y) = Q(y,x)$.

What he means by "intrinsically defined" is precisely that the object is independent of a choice of frame field or coordinate system. You know that $Q$ is intrinsically defined, so that ${\rm Ric}_p(x) = Q(x,x)/(n-1)$ is also intrinsically defined (you just found an expression that does not depend on the frame field).