# Is $\hat t$ orthogonal to the tangent space of the spatial hypersurface?

For a 3-D manifold ($$r,\theta,\phi$$) and sub-manifold (2-sphere)($$\theta,\phi$$) , $$\hat r$$ is orthogonal to both the 2-sphere and tangent space associated with each point of the 2-sphere. Similarly for 4-D manifold($$r,\theta,\phi,t$$) ,and sub-manifold(hypersurface)($$r,\theta,\phi$$) if $$\hat t$$ is orthogonal to the hypersurface(space like) for which the metric becomes $$ds^2=g(t)dt^2+f(t)\delta_{ij}\gamma_{ij}dx^idx^j$$(no cross terms between $$t$$ and $$x^{is}$$)where $$x^i$$ are the spatial coordinates ,is it necessary that $$\hat t$$ is also orthogonal to the tangent space associated with each point of the hypersurface??

The components of the metric tensor are defined as $$g_{\mu \nu} = \partial_\mu \cdot \partial_\nu$$, where $$\cdot$$ is the scalar product and $$\partial_\mu$$ is a coordinate basis vector.
• Yes it is but my question is if $\hat t$ is orthogonal to spatial hypersurface ,then whether $\hat t$ will also be orthogonal to tangent space associated with each point of the spatial hypersurface?? Mar 4, 2019 at 18:25
• The tangent space to each point of the spacelike hypersurface is made up of the coordinate bases associated to the coordinates $(r, \theta, \phi)$. A vector, in this case $\hat t$, is orthogonal to a hypersurface, if it is orthogonal to the tangent space of the hypersurface. It is a definition. Mar 5, 2019 at 8:38