# CR-Invariant powers of the sub-Laplacian-I

Today I will be reviewing the paper “CR-invariant powers of the sub-Laplacian” by Rod Gover and Robin Graham in order to hopefully understand it better. I will post images from the original paper, and then writing an explanatory commentary below.

Why do we care about powers of laplacians? Because they naturally comes up when we talk about obstructions to harmonic extensions, as can be seen in the GJMS paper. For each $k$, the weight of the density has to be $\frac{n}{2}+k$. This gives us the obstruction to harmonic extension.

How is the Fefferman metric relevant here though? Well if we recall the GJMS paper, it constructs these powers of the laplacian based on the Fefferman metric itself. That is how it becomes relevant. This is of course the Fefferman-Graham metric, that the co-author of this paper modestly refuses to put his own name to.

What does CR invariant mean? This will be explained at greater length below.

Why are invariant differential operators important? This is the wiki article on invariant differential operators. I would like to discuss some examples from this article.

1. Why does $\nabla$ have to be rotationally invariant?
2. The exterior derivative $d$ changes its representation with coordinates. What does it mean to be invariant? The commutative property with respect to group transformation is being referred to here.

Why is $k\leq N/2$ (for $N$ even) an important requirement? Because the $k$th coefficient of $\tilde{g}$ will be needed to define $\Delta^k$, and $\tilde{g}$ is not uniquely defined for $k>N/2$.

The conformal structure is on a circle bundle, and not a cone. I’m not sure. I suppose we will have to refer to the original paper to get an idea of why we have a circle bundle and not a cone.

What is the root of a bundle? An $n$th root of a bundle $X$ is a bundle $Y$ such that $Y^{\otimes^n}=X$. For instance, $\Bbb{R}_{\Bbb{R}}$ and $i\Bbb{R}_{\Bbb{R}}$ are square roots of $\Bbb{R}$.

Why do we need densities anyway? Why do we need homogeneous (in $t$) extensions of functions? I think that we are pre-empting how mathematical objects will change with a change in metric. But functions don’t change their value when there’s a change in metric! I think that we might be preserving a scalar property of these functions….like the integral. Hence, when we are dealing with “conformal quantities” that may change with a conformal change in metric, we want to preserve scalar quantities that might become observables (in Physics). Hence, we scale other quantities appropriately with the metric.

Why do we care about the operators $P_k$? They are just supposed to be powers of the Laplacian in the appropriate sense, which can be recovered from the obstruction tensor right? Well it’s technically $latex tf_{g}(\tilde{\Delta}^k_{TM})$, but yes, it is close enough. Note that $\Delta^k$ has nothing to do with $P_{k}$. We are concerned only with powers of the ambient laplacian. $\tilde{\Delta}^k|_{G}$ is independent of the extension of $\tilde{f}$ that is chosen, and $\tilde{\Delta}$ has an obstruction at order $Q^{k-1}$ such that $Q^{1-k}\tilde{\Delta}|_{G}$ is independent of the extension of $\tilde{f}$ mod $Q^{k}$. What about dependance on $\tilde{g}$? We’ve already taken care of that by choosing $k$ appropriately. There is no dependence on the chosen extension of $\tilde{g}$. Why is it important not to have dependence on the chosen extension of $f$? We are saying, in some sense, that we are defining data on the manifold in the form of $g$ and $f$, and we don’t want to make any other choices. And only with this data on the manifold, we should be able to define well-defined conformal invariants. Hence, it is important to construct quantities that don’t depend on the extensions of $\tilde{g}$ and $\tilde{f}$.

How can $\tilde{\Delta}^k|_{G}$ and $Q^{1-k}\tilde{\Delta}|_{G}$ ever be related? In some sense, the former is a multiple of the the $(k-1)$th order term of the latter’s Taylor expansion in terms of $Q$. However, it is important to note that this is just a heuristic explanation, and that $\tilde{\Delta}^k$ and $\tilde{\Delta}$ agree only on $G$.

Although the introduction continues after this, we will now jump to the next section of the paper.

Why do we study CR structures? Even dimensional manifolds can easily be complexified. However, this is difficult to do with odd dimensional manifolds. CR structures are the next best thing: odd dimensional manifolds can now be studied as hyper surfaces in larger complex manifolds. This allows us to import the heavy duty machinery of complex analysis in order to study these manifolds, perhaps study some invariant properties of theirs, and consequently classify them.

Why do we deal with co-vectors like $\Lambda$, and not just the tangent sub bundles? This is because differential forms lend themselves easily to cohomology, while vector fields don’t. Hence, we do so in order to use the machinery of cohomology.

What does $\Delta^{1,0}$ look like? One may imagine it to be $n+1$ complex dimensional, as it is “orthogonal” only to $\overline{L}$. What does a complex weight mean? I suppose that it is a formal object that lends itself to manipulation in this paper (this is probably the most vacuous sentence I’ve written in the recent past).

Let us take a small detour into CR manifolds for a bit. We will be discussing the wikipedia article on these manifolds.

Why do we complexify tangent spaces? This is because complexification is the only way that allows us to define holomorphic forms on the manifold. We also need to complexify the tangent bundles of complex spaces in order to define holomorphic forms on them. the complexification need not mean anything geometrically. It is just a formal tool.

Why do we need preferred distributions? Preferred distributions span the holomorphic vector fields that can be defined on the manifold. How do we know which vector fields are holomorphic? In the simple example of the manifold being embedded inside a larger complex manifold, the answer to this question is clear. The restrictions of holomorphic vector fields to the CR manifold form the preferred distribution $L$. Note that $L$ also needs to be integrable because we want the manifold $M$ to be a leaf, in case the larger complex manifold were foliated with respect to the sub bundle $L$ (assuming of course that $L$ is defined over the whole complex manifold, and not just on $M$.

What is important to note here is that the $L$ sub bundle of $S^3\subset \Bbb{C}^2$ does not consist of all possible holomorphic forms. It only consists of those forms that are “tangent” to it. But how can complex vector fields be tangent to $S^3$? Aren’t all vector fields over real numbers? We define tangency not by a visual picture of asymptotically touching a manifold, but algebraically. By that definition, the complex vector field mentioned above is tangent to $S^3$, ie it lies in the complexified tangent space of $S^3$. It’s not that only the vectors in the real tangent space of $S^3$ are “really” tangent to it. Tangency is now fair game for all vectors, real or complexified.

But now that we’re considering both real and complexified tangent vectors of $S^3$, won’t that double the dimension of the space of tangent vectors? Yes and no. Yes, over the reals, and no over $\Bbb{C}$. Also, note that $L$ here is just one (complex) dimensional, and consequently so is $\overline{L}$. Hence, $L\cup \overline{L}$ don’t span the complexified tangent bundle of the manifold. There is always an extra complex dimension left.

How can it be a metric on $L$ when it is a $2$-form? Well we choose two vectors in $L$, and then make one of them anti-holomorphic. How can we form a metric on a sub bundle of the (complexified) tangent bundle though? Shouldn’t it have been defined over the whole tangent bundle? That needn’t be the case. Metrics are mostly use to measure volumes. If all our sub manifolds will only have these holomorphic vector fields spanning their tangent spaces, we might as well define a metric only on this sub bundle. However, it is true that this metric will fail to give us the volume form of the whole space.

By focusing only on $L$, are we missing out on the geometry of the whole of $M$? Well we are yet to meet the Levi form. That will pick up the slack.

Why do we want $2$-forms to take values in vector bundles? Why not just $\Bbb{R}$ or $\Bbb{C}$? Because we want to generalize the notion of a function to a section of a vector bundle. Hence, it is only natural that we have forms that takes values in that vector bundle.

The Levi form maps pairs of vectors in $L$ to the complement of $L\oplus \overline{L}$. In a sense, we can construct the whole complexified tangent bundle using just $L$ and $h$, where $h$ is the Levi form.

Let us now get back to the paper.

What does the real part of a vector space mean? Real coefficients? Yes. Take all the coefficients, and consider only their real part. For instance, the real part of $\frac{\partial}{\partial z}_{\Bbb{C}}$ is $\{ \frac{\partial}{\partial z},\frac{\partial}{\partial y}\}_{\Bbb{R}}$. In practice, it is found by $z\to \frac{z+\overline{z}}{2}$.

What is happening with the dimensions? We start with a $2n+1$ dimensional manifold, perhaps embedded inside a $2n+2$ real dimensional space. After we complexify the tangent bundle of the space, we get a $2n+2$ complex dimensional space, in which $L$ is $n$ (complex) dimensional. The real part of $L$ has real dimension $2n$ (and similarly the complex part also has real dimension $2n$). Hence, having complex dimension $1$ over $\Bbb{C}$ is equivalent to having real dimension $4$ over $\Bbb{R}$. The complexification of the $2n+2$ real dimensional space consequently gives us a $8n+8$ real dimensional space over $\Bbb{R}$.

Is $H$ closed under the action of $J$? Yes. Although $L$ might not be closed under $J$, $H$ certainly is. How does that impose an orientation on $H$? Well it imposes an orientation on pairs of vectors, and that is all that is needed to impose an orientation on the whole even dimensional $H$.

What is the contact form exactly? It is a form (not a vector) which is orthogonal to $H$. It is part of $T^*M$ (without need of complexification), and there are multiple such choices possible (scalar multiples, etc). Where does $d\theta$ lie? Both $\theta$ and $d\theta$ can be written in terms of real coefficients and $\{dx^i,dy^j\}$. However, clearly $2i d\theta\in \Bbb{C}T^*M\wedge T^*M$. Its action is restricted to $T^{1,0}$. Moreover, $T$ is a vector field in the complexified tangent bundle such that $\theta(T)=1$ but $d\theta(T)=0$. Seeing as $theta$ is orthogonal to $H$ (and hence $T^{1,0}$), $T$ can have a maximum of $n+1$ unknown coefficients, and that’s exactly the number of equations we get from above. Hence, we can determine $T$ uniquely. Again note that $T$ is a complex vector field, and may not lie inside $TM$.

$\theta^{\alpha}$ are again in $\Bbb{C}T^*M$. Why couldn’t we just have defined them as generating the dual space of $T^{1,0}$? Because then they would not have been in the dual space of $T^{0,1}$, which they are.

How do we get that formula for $d\theta$? $-2id\theta$ gives us $2h_{\alpha\overline{\beta}}\theta^{\alpha}\wedge \theta^{\overline{\beta}}$. This does give us a restriction to $T^{1,0}$, although the fact of $2$ looks arbitrary.

What does the last formula mean? $T$ is outside of $T^{1,0}$. $\zeta$ is an $n+1$ form. Hence, the two contractions given on the right hand side give us $n$ forms in $T^{1,0}$ and $T^{0,1}$ respectively. Remember that we had a free choice of $\theta$. Hence, we can impose this restriction on the pseudohermitian form in order to get this succinct volume form (based on a prior choice of $\zeta$, of course).

What does “depends only on CR structure” mean? It means that the pseudohermitian form does not depend on $\lambda$. That is clearly the case with this definition. What does $\mathcal{E}(1,1)$ mean? It is a vector bundle of the form $\mathcal{E}(1)\otimes \overline{\mathcal{E}(1)}$. Isn’t the bold $\theta$ still just a section of $T^*M$? No, because the action of $\Bbb{R}$ on $|\zeta|^{-2/n+2}$ is different from the action of $\Bbb{R}$ on $\Bbb{R}$.

Why do we need $\zeta$ at all to determine a unique pseudohermitian form? Why can’t we just choose a form of unit length or something? We can’t because we don’t have any metric to measure lengths in $TM\setminus H$. Hence, we have the choice of a CR scale.

What is so “flat” about the Heisenberg group? Well, the Heisenberg group can be embedded as $\Bbb{R}^3\subset \Bbb{C}^2$. Hence, it is flat, as opposed to $S^3\subset \Bbb{C}^2$.

Do we have enough equations to determine $\omega_\alpha^\beta$? I suppose the torsion tensor will be given to us. We have $n^2$ variables and $n$ equations. I suppose we only need to determine $\omega_\alpha^\beta\otimes \theta^\alpha$, and not the individual $\omega$‘s? I’m not sure. I’m now confident that the torsion data has to be supplied beforehand, however.

Why are connections defined on the basis of $\omega$‘s? Well we can’t have Christoffel symbols because we don’t have a Riemannian metric. Hence, the hermitian metric $d\theta$ is the best we can do. But why do we not have a regular coordinate derivative, like in the definition of the Levi-Civita connection? Well we are only defining the connections with respect to the “coordinate co-vectors”. Maybe the partial derivatives will come in when we have more general co-vectors? The $\omega$‘s are pretty similar to the Christoffel symbols we have, with one index missing because we’re not really “parallel-transporting” in any direction. Also note that $\nabla h=0$.

What is the point of defining a connection, if we can already differentiate? The main point here is invariance. We don’t want derivatives to change when coordinate systems (observers, frames of reference, etc) are changed. Coordinate derivatives are not invariant. However, connection derivatives are. A similar notion is now being proposed for CR manifolds. Also, manifolds don’t always have a global coordinate system. Derivatives have to be defined “intrinsically” without reference to coordinates. Connections don’t need coordinates to be defined. They don’t depend on the intrinsic information of the manifold, which in the case of Riemannian manifolds is the metric.

This is all for today. I will try to upload the second part of this expository series tomorrow.

Graduate student