# Projection tensor in from (m+n) dim down on n-dim

1. Nov 28, 2013

### center o bass

Suppose that we have an (n+m)-dimensional tangent space $T_p^{n+m}$ which we decompose into the direct sum of two tangent spaces $T_p^{n+m} = T_p^n \oplus T_p^m$. We have a coordinate basis in some region of the manifold $\left\{\partial_{\mu}\right\}_{\mu=1}^{n+m}$ from which we want to construct an adopted basis $\left\{e_a, e_k\right\}$ such that $e_a \in T_p^{n}$ and $e_k \in T_p^m$ with $a = 1,2,\ldots , n$ and $k = 1, 2, \ldots m$. Now if $m=1$ one can easily construct a projection operator which projects down on the $n$-dimensional tangent-space $T_p^n$ by

$$P = \mathbb{1} - \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}}$$

where I have defined $e_k \equiv e_{n+1}$ and the metric-dual $e^k = g(e_k, \ )$, where $g$ is the metric tensor and $\mathbb{1}$ is the identity tensor. As a projection tensor must, we have that $P$ satisfies

$$P P = \mathbb{1} - 2 \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}} + \frac{1}{g^2_{(n+1)(n+1)}} e_{n+1} \otimes e^{n+1} e^{n+1}(e_{n+1}) = \mathbb{1} - \frac{e_{n+1}\otimes e^{n+1}}{g_{(n+1)(n+1)}} = P.$$

On the other hand, when $m > 1$ the similar expression

$P = \mathbb{1} - \sum_{k=1}^m\frac{e_{k}\otimes e^{k}}{g_{kk}}$

does not automatically satisfy $PP = P$, since

$$PP = \mathbb{1} - 2 \sum_{k=1}^m\frac{e_{k}\otimes e^{k}}{g_{kk}} + \sum_k \sum_l \frac{g_{kl}}{g_{kk}g_{ll}}e_k \otimes e^l.$$

So it's only a projection tensor if $g_{kl} = g_{kk} \delta_{kl}$ no, sum. Is there a general formula for the projection tensor down on $T_p^n$ that satisfies the required relations and for which we need not have $g_{kl} = g_{kk}\delta_{kl}$?