# Ngô Quốc Anh

## December 19, 2012

### How to decompose tensors into a purely spatial part and a timelike part?

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 15:02

Today, we aim to talk about how to decompose tensors into a purely spatial part which lies in the hypersurfaces $M$ and a timelike part which is normal to the spatial surface $M$?

Let us recall that $M$ (called spatial surface) is a hypersurface of $V$ (called the spacetime) of the dimension $n+1$. At each point $p\in M$, the space of all spacetime vectors can be orthogonally decomposed as $\displaystyle T_p(V)=T_p(M) \oplus \text{span}(n),$

where $\text{span}(n)$ stands for the 1-dimensional subspace of $T_p(V)$ generated by the unit normal vector $n$ to the surface $M$.

To do so, we need two projection operators.

The orthogonal projector onto $M$. In the literature, there exists such an operator, denoted by the symbol $\gamma$, given by $\displaystyle \begin{gathered} \gamma :{T_p}(V) \to {T_p}(M) \hfill \\ \qquad \quad\,\,\,v \mapsto v + (n \cdot v)n. \hfill \\ \end{gathered}$

According to the above decomposition and thanks to $\displaystyle g_V = g_M + n \otimes n$

with respect to any basis $(e_i)$ of the space $T_p(V)$, we have $\displaystyle \gamma_{ij}=g_{ij}+n_in_j,$

which, by raising indices, gives $\displaystyle \gamma^i_{j}=\delta^i_{j}+n^in_j.$