# Ngô Quốc Anh

## June 19, 2013

### The Bochner formula for vector fields

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 2:50

The Bochner formula for a gradient vector field is a important tool in geometric analysis which basically says that $\displaystyle\frac{1}{2}\Delta (|\nabla u{|^2}) = |{\nabla ^2}u{|^2} + \langle \nabla \Delta u,\nabla u\rangle + {\rm Ric}(\nabla u,\nabla u)$

for any function $f$. Whenever $|\nabla u| \ne 0$, we have a  formula for $|\nabla u|$ as follows $\displaystyle\frac{1}{2}\Delta (|\nabla u|) = \frac{1}{|\nabla u|} \left( \nabla u \cdot \nabla u ( \Delta u) + {\rm Ric}(\nabla u,\nabla u) + |\nabla \nabla u|^2 - \bigg| \Big\langle \nabla \nabla u, \frac{\nabla u}{|\nabla u|}\Big\rangle\bigg|\right).$

Today, we discuss a variant of it called the Bochner formula for vector fields. I found this identity in a recent preprint of Li Ma [here].

Theorem. Let $(M, g)$ be a Riemannian manifold of dimension $n$. Let $X$ be a smooth vector field on $M$. Then we have $\displaystyle\frac{1}{2}\Delta (|X|^2) = |\nabla X|^2 + {\rm div}({\mathbb L_X}g)(X) - {\nabla _X}{\rm div}X - {\rm Ric}(X,X)$

where ${\mathbb L_X}g$ is the Lie derivative of the vector field $X$ with respect to underlying metric $g$.

## June 14, 2013

### MuPAD: Drawing asymptotically flat spacetime

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 23:06

Today, I will show how to draw an asymptotically flat spacetime with two ends using MuPad. To draw ends, we use hyperbolic functions.

First, we talk about the end. The function that I am going to use is $h(t)=\frac{1-4t}{t-2}+12$. It is clear that $h$ blow up at $t=2$ and approaches $-4$ at infinity. To use $h$, we use the following

h := proc(t)
begin
(1-4*t)/(t-2)+12
end_proc

I have added $12$ to the function $h$ so that $h$ approaches $8$ at infinity. Then to draw the (upper) end, I use the following function

f := proc(x, y)
begin
if
x^2 + y^2 > 3.4
then
h(x^2+y^2)
else
end_if
end_proc

I have used the number $3.4$ because I do not want my end is too tall, keep in mind that $2$ is the blow-up number. We can calculate

h(3.4)

to see that this number is nothing but $3$ which is close to $1$ as I need. We are now able to draw the first end using the following

plot(
plot::Function3d(f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
ViewingBox = [Automatic, Automatic, 3 .. 8],
Scaling = Constrained)