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(adsbygoogle = window.adsbygoogle || []).push({});

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

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}##?

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

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