# Ngô Quốc Anh

## November 26, 2009

### R-G: Scalar curvature

Filed under: Riemannian geometry — Ngô Quốc Anh @ 20:51

In Riemannian geometry, the scalar curvature (or Ricci scalar) is the simplest curvature invariant of a Riemannian manifold. To each point on a Riemannian manifold, it assigns a single real number determined by the intrinsic geometry of the manifold near that point. Specifically, the scalar curvature represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. In two dimensions, the scalar curvature is twice the Gaussian curvature, and completely characterizes the curvature of a surface. In more than two dimensions, however, the curvature of Riemannian manifolds involves more than one functionally independent quantity.

In general relativity, the scalar curvature is the Lagrangian density for the Einstein–Hilbert action. The Euler–Lagrange equations for this Lagrangian under variations in the metric constitute the vacuum Einstein field equations, and the stationary metrics are known as Einstein metrics. The scalar curvature is defined as the trace of the Ricci tensor, and it can be characterized as a multiple of the average of the sectional curvatures at a point. Unlike the Ricci tensor and sectional curvature, however, global results involving only the scalar curvature are extremely subtle and difficult. One of the few is the positive mass theorem of Richard Schoen, Shing-Tung Yau and Edward Witten. Another is the Yamabe problem, which seeks extremal metrics in a given conformal class for which the scalar curvature is constant.

Definition. The scalar curvature is the function $S$ defined as the trace of the Ricci tensor.

Since the Ricci tensor is an $(2,0)$-tensor field then in the local coordinates

$S = {\rm Trace}( {\rm Ric}) = g^{jk}R_{jk}$.

Theorem (Contracted Bianchi Identity). The covariant derivatives of the Ricci and scalar curvatures satisfy the following identity

$\displaystyle {\rm div} {\rm Ric} = \frac{1}{2} \nabla S$.

Examples 1. We still work on the two-dimensional spherical surface of radius $R$ whose metric is

$\displaystyle \left( {{g_{ij}}} \right) = {R^2}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {{{\sin }^2}\theta } \\ \end{array} } \right)$

as the previous topic. Then

$\displaystyle S = {g^{jk}}{R_{jk}} = {g^{11}}{R_{11}} + {g^{22}}{R_{22}} = \frac{2}{{{R^2}}}$.

Examples 2. We now work for the two-dimensional space-like “upper hyperboloid” of the Minkowski space whose metric is

$\displaystyle {\left( {ds} \right)^2} = \frac{{{R^2}}}{{{r^2} + {R^2}}}{\left( {dr} \right)^2} + {r^2}{\left( {d\phi } \right)^2}$

that is

$\displaystyle \left( {{g_{ij}}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{{{R^2}}}{{{r^2} + {R^2}}}} & 0 \\ 0 & {{r^2}} \\ \end{array} } \right),\left( {{g^{ij}}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{{{r^2} + {R^2}}}{{{r^2}}}} & 0 \\ 0 & {\frac{1}{{{r^2}}}} \\ \end{array} } \right)$.

Then

$\displaystyle {R_{11}} = - \frac{1}{{{r^2} + {R^2}}},{R_{12}} = {R_{21}} = 0,{R_{22}} = - \frac{{{r^2}}}{{{R^2}}}$

and

$\displaystyle S = - \frac{2}{{{R^2}}}$.

### R-G: Ricci curvature

Filed under: Riemannian geometry — Ngô Quốc Anh @ 19:56

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, represents the amount by which the volume element of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space. As such, it provides one way of measuring the degree to which the geometry determined by a given Riemannian metric might differ from that of ordinary Euclidean n-space. More generally, the Ricci tensor is defined on any pseudo-Riemannian manifold. Like the metric itself, the Ricci tensor is a symmetric bilinear form on the tangent space of the manifold.

The Ricci curvature is broadly applicable to modern Riemannian geometry and general relativity theory. In connection with the latter, it is up to an overall trace term, the portion of the Einstein field equation representing the geometry of spacetime, the other significant portion of which comes from the presence of matter and energy. In connection with the former, lower bounds on the Ricci tensor on a Riemannian manifold allow one to extract global geometric and topological information by comparison (cf. comparison theorem) with the geometry of a constant curvature space form. If the Ricci tensor satisfies the vacuum Einstein equation, then the manifold is an Einstein manifold, which have been extensively studied (cf. Besse 1987). In this connection, the Ricci flow equation governs the evolution of a given metric to an Einstein metric, the precise manner in which this occurs ultimately leads to the solution of the Poincaré conjecture.

Definition. Ricci curvature (or Ricci tensor) is an $(2,0)$-tensor field denoted by $\rm Ric$, that is ${\rm Ric} : TM \times TM \to \mathbb R$, defined by

${\rm Ric}(X,Y) = {\rm Trace}( x \to R(x, X)Y)$.

In local coordinates, $\rm Ric$ is of the form

${\rm Ric} = R_{ij} dx^i \otimes dx^j$.

We assume $\frac{\partial}{\partial x^i}$ where $i=1,2,...,n$  is an orthonormal basis for $T_pM$, then

$\displaystyle R\left( {\frac{\partial }{{\partial {x^i}}},X} \right)Y = {X^j}{Y^k}R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^k}}} = {X^j}{Y^k}R_{kij}^l\frac{\partial }{{\partial {x^l}}}$.

Thus,

$\displaystyle {\rm Trace}\left( {x \mapsto R\left( {x,X} \right)Y} \right) = {X^j}{Y^k}R_{kij}^i$.

In other words,

$\displaystyle {\rm Ric}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{ij}}d{x^i} \otimes d{x^j}\left( {\frac{\partial }{{\partial {x^j}}},\frac{\partial }{{\partial {x^k}}}} \right) = {R_{jk}} = R_{kij}^i$.

To be exact, one should read

$\displaystyle {R_{jk}} = \sum\limits_i {R_{jik}^i}$.

A simple calculation shows us that

$\displaystyle {R_{jk}} = R_{jik}^i = {g^{li}}{g_{li}}R_{jik}^i = {g^{li}}{R_{ljik}}$.

Thus, Ricci tensor can be thought as the trace of curvature tensor $R_{ljil}$.

Example. For the two-dimensional spherical surface of radius $R$ whose metric is

$\displaystyle{\left( {ds} \right)^2} = {R^2}\left[ {{{\left( {d\theta } \right)}^2} + {{\sin }^2}\theta {{\left( {d\phi } \right)}^2}} \right]$

we have

$\displaystyle \left( {{g_{ij}}} \right) = {R^2}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {{{\sin }^2}\theta } \\ \end{array} } \right), \qquad \left( {{g^{ij}}} \right) = \frac{1}{{{R^2}}}\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {\frac{1}{{{{\sin }^2}\theta }}} \\ \end{array} } \right)$.

Thus

$\displaystyle\begin{gathered} {R_{11}} = R_{111}^1 + R_{121}^2 = {g^{22}}{R_{2121}} = {g^{22}}{R_{1212}} = 1, \hfill \\ {R_{12}} = R_{112}^1 + R_{122}^2 = 0, \hfill \\ {R_{22}} = R_{212}^1 + R_{222}^2 = {g^{11}}{R_{1212}} = {\sin ^2}\theta , \hfill \\ {R_{21}} = {R_{12}} = 0. \hfill \\ \end{gathered}$

### R-G: Sectional curvature

Filed under: Riemannian geometry — Ngô Quốc Anh @ 14:40

In Riemannian geometry, the sectional curvature is one of the ways to describe the curvature of Riemannian manifolds.

Definition. The sectional curvature of the plane spanned by the (linearly independent) tangent vectors $X, Y \in T_xM$ of the Riemannian manifold $M$ is

$\displaystyle K\left( {X,Y} \right) = \frac{{\left\langle {R\left( {X,Y} \right)Y,X} \right\rangle }}{{\left\langle {X,X} \right\rangle \left\langle {Y,Y} \right\rangle - {{\left\langle {X,Y} \right\rangle }^2}}}$.

In local coordinates, if

$\displaystyle X = {X^i}\frac{\partial }{{\partial {x^i}}}, \quad Y = {Y^j}\frac{\partial }{{\partial {x^j}}}$

we then have

$\displaystyle R\left( {X,Y} \right)Y = {X^i}{Y^j}{Y^k}R\left( {\frac{\partial }{{\partial {x^i}}},\frac{\partial }{{\partial {x^j}}}} \right)\frac{\partial }{{\partial {x^k}}} = {X^i}{Y^j}{Y^k}R_{kij}^l\frac{\partial }{{\partial {x^l}}}$

which implies

$\displaystyle\begin{gathered} \left\langle {R\left( {X,Y} \right)Y,X} \right\rangle = {X^i}{Y^j}{Y^k}R_{kij}^l\left\langle {\frac{\partial }{{\partial {x^l}}},{X^m}\frac{\partial }{{\partial {x^m}}}} \right\rangle \hfill \\ \qquad= {X^i}{Y^j}{X^m}{Y^k}R_{kij}^l{g_{lm}} \hfill \\ \qquad= {R_{mkij}}{X^i}{Y^j}{X^m}{Y^k} \hfill \\ \qquad = {R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}. \hfill \\ \end{gathered}$

Besides

$\displaystyle\begin{gathered} \left\langle {X,X} \right\rangle \left\langle {Y,Y} \right\rangle - {\left\langle {X,Y} \right\rangle ^2} = {X^i}{X^m}{g_{im}}{Y^j}{Y^k}{g_{jk}} - {\left( {{X^\alpha }{Y^\beta }{g_{\alpha \beta }}} \right)^2} \hfill \\ \qquad= {X^i}{X^m}{g_{im}}{Y^j}{Y^k}{g_{jk}} - {X^\alpha }{Y^\beta }{g_{\alpha \beta }}{X^\gamma }{Y^\delta }{g_{\gamma \delta }} \hfill \\ \qquad= \left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}. \hfill \\\end{gathered}$

Thus

$\displaystyle K\left( {X,Y} \right) = \frac{{{R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}}}{{\left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}}}$.

To be exact, without using Einstein summation convention, one reads the above identity as following

$\displaystyle K\left( {X,Y} \right) = \frac{{\sum\limits_{ijmk} {{R_{ijmk}}{X^i}{Y^j}{X^m}{Y^k}} }}{{\sum\limits_{ijmk} {\left( {{g_{im}}{g_{jk}} - {g_{ij}}{g_{mk}}} \right){X^i}{X^m}{Y^j}{Y^k}} }}$.

We refer the reader to this topic for examples. In addition, if we choose

$\displaystyle {g_{ij}} = {\left( {\displaystyle\frac{2}{{1 - {{\left| y \right|}^2}}}} \right)^2}{\delta _{ij}}$

then the sectional curvature of $g$ is $-1$.

### R-G: Hessian and Laplacian

Filed under: Riemannian geometry — Ngô Quốc Anh @ 1:17

For a given smooth function $f$ on manifold $M$, the gradient of $f$ is given by

$\displaystyle \nabla f = g^{kj} \dfrac{\partial f}{\partial x^j} \frac{\partial}{\partial x^k}$.

Note that gradient of $f$ is also a vector field on $M$. Thus, for each $X \in TM$, it is reasonable to talk about $\nabla_X \nabla f$.

Definition 1. Hessian of $f$, denoted by ${\rm Hess}$, is defined as the symmetric $(0,2)$-tensor

${\rm Hess} f (X,Y)=g(\nabla_X \nabla f, Y)$.

We also denote by $f_{ij}$ the following

$\displaystyle {\rm Hess} f \left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right)$.

Thus

$\displaystyle\begin{gathered} {f_{ij}} = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\nabla f,\frac{\partial }{{\partial {x^j}}}} \right) = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}\frac{\partial }{{\partial {x^k}}}} \right),\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = g\left( {\frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^k}}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)g\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\\quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}} \left[ \frac{1}{2}\left( {-\frac{{\partial {g_{ki}}}}{{\partial {x^j}}} + \frac{{\partial {g_{ij}}}}{{\partial {x^k}}} + \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}} \right)\right]. \hfill\end{gathered}$

Note that

$\displaystyle \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + {g^{kl}}\frac{\partial }{{\partial {x^i}}}\left( {\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}}$.

Since $0=\frac{\partial}{\partial x^i}(g^{kl}g_{kj})$ then

$\displaystyle\frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} = - \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g^{kl}}$

which implies

$\displaystyle f_{ij} =\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^m\frac{{\partial f}}{{\partial {x^m}}}$.

Definition 2. Laplacian of $f$, denoted by $\Delta f$, is defined as the trace of ${\rm Hess} f$.

Note that ${\rm Hess} f$ is a $(0,2)$-tensor, then in local coordinates, one has

$\displaystyle \Delta f = {g^{ij}}{f_{ij}}$.

It is clear that $\nabla X$ is a $(1,1)$-tensor field. To see this fact, one can assume $X=X^i \frac{\partial}{\partial x^i}$ then from

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^j}}}}}\left( {{X^i}\frac{\partial }{{\partial {x^i}}}} \right) = \frac{{\partial {X^i}}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}} + {X^i}\Gamma _{ji}^l\frac{\partial }{{\partial {x^l}}}$

one has

$\displaystyle\nabla X = \left[ {\frac{{\partial {X^i}}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}} + {X^i}\Gamma _{ji}^l\frac{\partial }{{\partial {x^l}}}} \right] \otimes d{x^j}$

since ${\nabla _Y}X = \left\langle {Y,\nabla X} \right\rangle$ which is exactly an $(1,1)$-tensor. Then we can define divergence of a vector field $X$ as following

Definition 3. Divergence of vector field $X$ is given by

$\displaystyle {\rm div} X = {\rm Trace}(\nabla X)$.

In coordinates, this is

$\displaystyle {\rm div} X = dx^i \left( \nabla_{\frac{\partial}{\partial x^i}} X\right)$

and with respect to an orthornormal basis

$\displaystyle {\rm div} X =g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,\frac{\partial }{{\partial {x^i}}}} \right)$.

Thus $\Delta f = {\rm Trace}(\nabla(\nabla f)) = {\rm div}(\nabla f)$.

NOTICE: To avoid any inconvenience caused, from now we denote gradient of $f$ by ${\rm grad}f$ instead of $\nabla f$. This is because $\nabla f$ is covariant derivative of $f$, this is an $(1,0)$-tensor instead of a vector field as mentioned in this entry.