I will use a local coordinate formalism here, since this is related to research in general relativity, and my supervisor only tolerates local coordinate formalisms. Plus the research papers I base my research on also use local coordinates.


If $M$ is an $n$ dimensional $C^\infty$ manifold, with local coordinates $(x^\mu)=(x^1,...,x^n)$, and $F:\Sigma\rightarrow M$ is a $C^\infty$ embedding of an $n-1$ dimensional smooth manifold $\Sigma$ into $M$, where $\Sigma$ is described by local coordinates $(\xi^i)=(\xi^1,...,\xi^{n-1})$, then "covariant" tensor fields $A\_{\mu\_1,...,\mu\_k}$ may be naturally pulled back to $\Sigma$ by $$ A\_{i\_1...i\_k}=e^{\mu\_1}\_{i\_1}...e^{\mu\_k}\_{i\_k}A\_{\mu\_1...\mu\_k}, $$ where $e^\mu\_i=\partial x^\mu/\partial \xi^i$ is both the matrix of the tangent map and the components of the (pushforward of the) coordinate frame $\partial/\partial\xi^i$ in the system $x^\mu$. Likewise, if $T^{i\_1...i\_k}$ is a "contravariant" tensor field on $\Sigma$, then it may be pushed forward to $M$ by $$ T^{\mu\_1...\mu\_k}=e^{\mu\_1}\_{i\_1}...e^{\mu\_k}\_{i\_k}T^{i\_1...i\_k}. $$


On the other hand, $F$ as an embedding is a diffeomorphism onto its image, thus it **should** be possible to pull back those contravariant tensor fields defined along the image of $\Sigma$ in $M$ that are "tangent" to $\Sigma$ where tangency can be defined as having vanishing contractions with the one-form $n\_\mu$, where the $n\_\mu$ is given by $n\_\mu\sim\partial\_\mu\phi$ where $\phi$ is a scalar field whose level set locally generates $F(\Sigma)$.


One possible construction I see to give this pullback explicitly is to construct a local foliation of $M$ of which (an open subset of) $F(\Sigma)$ is a leaf and set new coordinates $(x^{\mu'})=(\xi^1,...,\xi^{n-1},\phi)$, where the scalar field $\phi$ is fired up in such a way that the leaves of the foliation correspond to different level sets of $\phi$. Then, if $T^{\mu\_1...\mu\_k}$ is a contravariant tensor field along $F(\Sigma)$, the pullback $T^{i\_1...i\_k}$ can be defined as transforming to the slice chart $(x^{\mu'})$ and restricting the range of the the indices $\mu'$ to $1,...,n-1$, since the $\phi$-components will be zero anyways.




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With this said, I can present my problem. If a Lorentzian metric $g$ is given on $M$, and the induced metric on $\Sigma$, $\gamma=F^\*g$, $\gamma\_{ij}=e^\mu\_ie^\nu\_jg\_{\mu\nu}$ is nondegenerate, then the induced connection on $\Sigma$ is usually given as $$ D\_iT^{i\_1...i\_k}\_{j\_1...j\_l}=e^\mu\_i e\_{\mu\_1}^{i\_1}...e^{\nu\_1}\_{j\_1}...\nabla\_\mu T^{\mu\_1...\mu\_k}\_{\nu\_1...\nu\_l} ,$$ where greek indices are raised/lowered by $g$ and latin indices are raised/lowered by $\gamma$, and $T$ is *tangent* to the hypersurface. It can be easily seen that this connection on $\Sigma$ is precisely the Levi-Civita connection of $\gamma$.


If $\gamma$ is degenerate, then it is usually said that there is no unique induced connection on $\Sigma$, however the pullback of the covariant derivative on $M$ can be defined the same way as above, except all the latin indices will be covariant incides, because $\gamma^{ij}$ doesn't exist to raise indices.


So if we are given a tensor on $\Sigma$, we can express it in terms of greek indices (covariant latin indices can be made covariant greek indices the same way contravariant greeks can be made latin I detailed in the early part of my post), then we can calculate its covariant derivative in $M$, then project it down into $\Sigma$ with covariant indices, essentially the process is as such: $$ T^{i\_1...i\_k}\_{j\_1...j\_l}\rightarrow T^{\mu\_1...\mu\_k}\_{\nu\_1...\nu\_l}\rightarrow e^\mu\_i\nabla\_\mu T^{\mu\_1...\mu\_k}\_{\nu\_1...\nu\_l}\rightarrow D\_iT\_{i\_1...i\_kj\_1...j\_l}.$$


Admittedly, I don't know if it is possible to give any connection coefficients on $\Sigma$, which describes this process.


On the other hand, consider a contravariant vector field $A^\mu$ that is tangent to $\Sigma$. Then $$ D\_iA\_j=e^\mu\_i e\_{\nu j}\nabla\_\mu A^\nu=e^\mu\_i(\nabla\_\mu(e\_{\nu j}A^\nu)-A^\nu\nabla\_\mu e\_{\nu j})= \\ =\partial\_i A\_j-A^\nu e^\mu\_i\nabla\_\mu e\_{\nu j}=\partial\_i A\_j-A^ke^\nu\_k e^\mu\_i\nabla\_\mu e\_{\nu j}=\partial\_i A\_j-A^k\Gamma\_{k,ij}, $$ and I haven't checked but pretty sure the $\Gamma\_{k,ij}$s are given by the usual formula of the Christoffel symbols of the first kind for $\gamma\_{ij}$.


I guess, my question is, what can we say about induced covariant derivatives on null hypersurfaces? The general relativity literature seems to be adamant that it is ill-defined, but it seems to me it can be defined in certain cases. What further causes confusion in me is that it appears to me that one CAN in fact raise indices on $\Sigma$. Take $T\_{i\_1...i\_k}$ on $\Sigma$, push it forward to $T\_{\mu\_1...\mu\_k}$ by the procedure given in my post, then raise indices with $g^{\mu\nu}$, then pull back $T^{\mu\_1...\mu\_k}$ to $T^{i\_1...i\_k}$ once again with the process using the slice chart.


Thus, it seems to me that there actually **is** a way to define induced covariant derivatives on $\Sigma$, but obtaining it is not "nice". Is my conclusion correct? I am very confused, any clarification is much appreciated.



Induced connection on null hypersurfaces
The simple answer is: 


**You can't project to a null hypersurface**


This is tied to the fact that the "Lorentzian normal" vector field is in fact a tangent vector field. 




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In more details: 


Start with a tangent vector field $T^\nu$ to the image of $\Sigma$ in your Lorentzian manifold. If you lower the index by the metric $g\_{\mu\nu}$, what you get is no longer necessarily a "tangent one form". Therefore it no longer makes sense to "pull it back". 


An explicit example is taking the null foliation of $\mathbb{R}^{1,1}$. Let $u$, $v$ be functions such that the Minkowski metric looks like
$$ 2 \mathrm{d}u ~\mathrm{d}v $$
And consider the vector field $\partial\_u$ which is null, and is tangent to the level sets of $v$. Its metric dual is $2 \mathrm{d}v$ which restricts to zero (as a one form) on the level sets. 




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You should think that the induced connection in the non-degenerate setting really is saying the following:


The metric gives a canonical (unique) splitting of $TM$ along $\Sigma$ into $T \Sigma$ and $N\Sigma$, the normal bundle. Then the connection on $M$ induces a connection on $\Sigma$ by keeping only the part that is in $T \Sigma$ and tossing out everything that is in $N\Sigma$.


In the case $\Sigma$ is null, there is no unique splitting! In fact, $N\Sigma$ by the metric definition is part of $T\Sigma$! What you are missing is indeed just the projection: if you fix any choice of a transversal distribution to $T\Sigma$ in $TM$, then you are done: by linear algebra you can decompose any vectorfield in $TM$ into a portion in $T\Sigma$ and a portion in the transversal distributions. In the nondegenerate case the transversal distribution is provided by the normal bundle. In the degenerate case you don't have anything. (Which is why computations always fix, first and foremost, a null frame: once you fix a frame then you have a preferred transversal vector field and the theory goes through. But the choice of null frame is non-unique, which is why you have the entire theory of the GHP calculus.)




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To add a few more details:



> 
> If $\gamma$ is degenerate, then it is usually said that there is no unique induced connection on $Σ$, however the pullback of the covariant derivative on $M$ can be defined the same way as above, except all the latin indices will be covariant incides, because $\gamma^{ij}$ doesn't exist to raise indices.
> 
> 
> 


That is strictly not true. What you have defined is the the pull-back covariant derivative of the pullback of the cotangent bundle on $M$. This does not give you a covariant derivative on the cotangent bundle on $\Sigma$, since there is an ambiguity involved in how to "pushforward" $T^\*\Sigma$ to $T^\*M$. 


Revisiting the example with the two dimensional Minkowski space above, the issue is that given a one form tangent to $\Sigma$ being a level set of $v$, you can write it as, in the coordinate $u$, some function $f(u) ~\mathrm{d}u$. But you do not know how to canonically associate to this one form a one form on $M$: for any function $g(u,v)$, the one form $f(u) ~\mathrm{d}u + g(u,v) ~\mathrm{d}v$ restricts to your given one form on $\Sigma$. So knowing how to differentiate on the pull back of the cotangent bundle doesn't help you do anything with objects in $T^\*\Sigma$. 


The same mistake cropped up in 



> 
> So if we are given a tensor on $Σ$, we can express it in terms of greek indices (covariant latin indices can be made covariant greek indices the same way contravariant greeks can be made latin I detailed in the early part of my post)
> 
> 
> 


Contravariant tensors on $\Sigma$ can be expressed in terms of "greek indices", but covariant tensors cannot, because of the ambiguity described above. 


