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## Chern classes

Nothing new here.  Just exercises for myself.

If $E \rightarrow X$ is a holomorphic vector bundle on a complex manifold and $A$ is a connection then the curvature $F_A$ is a 2-form with coefficients in the $End(E)$.  The Chern class $c_{2k}(E)$ is given by the degree $2k$ closed form $\sigma_k( \frac{i}{2\pi} F_A )$ where $\sigma_k$ is the polynomial invariant under conjugation that associates the $k$-th symmetric function of the eigenvalues.  The Chern classes of a complex manifold is defined to be that of the tangent bundle.  For $\mathbf{CP}^1$ the hermitian metric is $h=\frac{dzd\bar{z}}{(1+|z|^2)^2}$ and $\Omega = 2 \frac{dz \wedge d\bar{z}}{(1+|z|^2)^2}$.  Recall the symmetric polynomials are 1, $x_1+x_2+\dots+x_n$, $\sum_{i and so on.  So the first Chern class will be the sum of the eigenvalues of the matrix $\frac{i}{2\pi} \Omega$ considered as a matrix $T^*\mathbf{CP}^1 \rightarrow T^* \mathbf{CP}^1$.  So $\int c_1 = \frac{i}{\pi} \int \frac{ dz \wedge d\bar{z} }{(1+|z|^2)^2} = \int dx dy/(1+x^2+y^2)^2$.  Evaluate $\int_{-\infty}^\infty dx/(a^2+x^2)^2$ by contour integration for fixed radius $R>0$ by drawing a circular arc $\Gamma_R$  around the origin which will include poles at $+ia,ia$ and the other component the $[-R,R]$ so $2\pi i Res( 1/(1+z^2)^2, -ia) + 2 \pi i Res( 1/(a+z^2)^2, +ia ) = \int_{-R}^R dx/(a^2 + x^2)^2 + \int_{\Gamma_R} dx/(a^2+x^2)^2$.  The second term is dominated by $\pi R/R^4$ which will go to zero.  So

$\lim_{R\rightarrow\infty} \int dx/(a+x^2)^2 = 2\pi i Res(1/(a+z^2)^2,-ia) + 2\pi i Res( 1/(a+z^2)^2, +ia)$  The poles are of order 4.  We have to use the formula $Res(f,c) = \frac{1}{(n-1)!}\lim_{z\rightarrow c} \frac{d^{n-1}}{dz^{n-1}} ((z-c)^n f(z)}$.

The residue at $z=ia$ is $\lim_{z\rightarrow ia} \frac{1}{3!} \frac{d^3}{dz^3} ( \frac{z-ia}{(z+ia)^2 )$  which is a messy computation I don’t feel like doing right now.

For the complex projective space, $\mathbf{CP}^n$ the total Chern class is $(1+a)^n$ where $a\in H^2(\mathbf{CP}^n)$ is $-c_1(\mathcal{()}(-1)} = c_1( \mathcal{O}(1)^*$.  The tautological line bundle sheaf, the holomorphic sections of the tautological line bunles is  $\mathcal{O}(-1)$.  If we realize $L$ a complex line through origin in $\mathbf{C}^{n+1}$ as a point of $\mathbf{CP}^n$ then the tautological line bundle $S$ is defined to be the sub-bundle of $\mathbf{CP}^n \times \mathbf{C}^n$ where at the point $[L]$ the fiber is the vector space $L$.  It’s holomorphic sections form the sheaf $\mathcal{O}(-1)$ by definition so then one defines $\mathcal{O}(1)$ as the dual sheaf and the various $\mathcal{O}(d)$ are $a$-th tensor powers of $\mathcal{O}(1)$.

$\mathbf{C}^n \rightarrow \mathbf{CP}^n$ is the standard projection.  For any complex line $L$ through the origin the tangent space $T \mathbf{CP}^n$ is identified with the linear functions from $L$ to its orthogonal complement.

There is an exact sequence $0 \rightarrow \mathcal{O}_{CP^n} \rightarrow \mathcal{O}(1)^{\oplus (n+1)} \rightarrow T \mathbf{CP}^n \rightarrow 0$.  I don’t understand the standard argument yet.  $T\mathbf{CP} = Hom( \mathcal{O}(-1), K)$ where $\mathcal{O}(1) \oplus K = \mathcal{ O }^{\oplus (n+1)}$.

Then $TCP^n \oplus \mathcal{O} = Hom( \mathcal{O}(-1),K) \oplus Hom( \mathcal{O}(-1), \mathcal{O}(-1))$.  By the Whitney formula

$c( \mathbf{CP}^n ) = c( T \mathbf{CP}^n ) = c( \mathcal{O}(1))^{n+1} = (1+a)^{n+1)$ where $a \in H^2(\mathbf{CP}^n,\mathbf{Z})$ is the generator of $H^2$ given by the negative of the first Chern class of the tautological line bundle $\mathcal{O}(-1)$ using $c_1(E^*) = -c_1(E)$.  So $c_k = (n,k) a^k$.

Now for something slightly more interesting from the point of basic curiosity.  What are the Chern classes of a projective variety defined by two of homogeneous equations of degrees $a,b$.  This is a complete intersection.  Its degree is $ab$.  Suppose $i:X \rightarrow \mathbf{CP}^n$ be the inclusion map.  We have $T\mathbf{CP}^n = TX \oplus N_X$ the tangent and normal decomposition.

$\langle i^* c_k(T\mathbf{CP}^n), [X] \rangle = deg(X) (n,k)$

Ok, so I just put in this note a link to an article with discussion of practical computations of Chern classes of algebraic varieties that I am putting down as homework for myself.  I don’t understand what Chern classes really mean geometrically and analytically.  And I guess my education on actually computing Chern classes is pretty weak, which is the point of this note.  In geometry the only examples of things are generally just algebraic objects and homogeneous objects and it’s a little sad that I don’t know how to compute Chern classes for these objects in this late stage of my life.

## Attempt to understand Donaldson-Sun theory

Years ago, when I was an undergraduate at Princeton and before I totally lost my way in mathematics, getting married, getting a job at Lehman,  and destroying my graduate studies at Columbia where John Morgan was kind enough to give me an opportunity and returning to Columbia having not studied Spin Geometry of Lawson-Michelson and instead seeking some crazy idea of not working on four manifolds (at the time it was the original $SU(2)$ theory) going to Dan Stroock and not delivering on the analysis that he expected and so on, I actually had a good intuitive understanding of some of the issues.  It’s a good idea for me to get a real education in mathematics, so here goes.

Step by step, slowly.

If $X$ is a complex Kahler manifold and $L$ is a line bundle and $A$ is a connection on the line bundle, then this defines a holomorphic structure on $L$.

Recall that a Kahler manifold $X$ has a hermitian metric $h$, i.e. at each point $p$ we have $h_p( \eta, \bar{\zeta}) = \bar{ h_p(\zeta, \bar{\eta})}$ and $h(\zeta,\bar{\zeta})>0$ and the 2-form $\omega_p(u,v) := h( iu, v)$ is closed satisfying $d\omega = 0$.  Basic properties of these objects and some known results are summarized here.  The complex structure $J: TX\otimes \mathbf{C} \rightarrow TX\otimes\mathbf{C}$ satisfied $J^2 = -1$ and therefore $TX\otimes\mathbf{C} = T^{0,1}M \oplus T^{1,0}M$ are the $+i,-i$-eigenspaces of $J$.  Using this decomposition one considers the sections of differential forms $\Gamma( \Wedge^k T^*M)$ which split up into $(p,q)$ forms.  The exterior derivative splits into $d = \partial + \bar{\partial}$ and the de Rham complex has a corresponding splitting.  Hodge theory gives $H^k(X) = \Oplus_{p+q=k} H^{(p,q)}(X)$.

A connection $A$ is equivalent to a map $\nabla_A: \Omega^0(X,E)\rightarrow\Omega^1(X,E)$ satisfying $\nabla_A(f\sigma) = f \nabla_A(\sigma) + df \otimes \sigma$.

Ok so in this basic setup, the holomorphic structure defined by the connection is  a map $\bar{\partial}_A: \Gamma(E)\rightarrow\Omega^{(0,1)}(X,E)$ defined by $\bar{\partial}_A (f \sigma) = f \bar{\partial}_E \sigma + \bar{\partial}f \otimes \sigma$.

So this is right, as we can check with some presentation of the basic issue.  I am going to be spending some time carefully picking up the pieces of the setting of interest in Donaldson-Sun theory because I am extremely rusty on these matters.

So Donaldson-Sun theory is taking a class of Kahler manifolds with bounds on volume and so on for which Gromov Compactness applies and then using Sobolev inequality and Hormander’s technique with UNIFORM CONTROL over this parameter space to get control of embedding into a complex projective space. Just as MOSER had looked at NASH embedding and picked out the softer scheme that took over mathematics, I am wondering whether it is a good project for ZULF to look at DONALDSON’s delicate work of genius and pull out a softer UNIVERSAL technique here. You take Gromov-Hausdorff convergence, you ask for what sort of things happen in the limit and control singularities of various types. Let’s remember this idea. This could change mathematics if implemented. WITTEN had looked at Donaldson’s earlier genius work and realized that softer Seiberg-Witten invariants can reprove and extend the topological results. In the same way, detailed constructions and estimates in his current work is likely to contain some universalizable things. It’s an obvious sort of thought and one that Donaldson himself might be looking for directly himself but sometimes there is value in taking vast distance.

## The co-area formula exercise from Terry Tao’s blog

This is an exercise from Terry’s blog on Hausdorff measures.

In order to unrustify my mathematics.

Implicit Function Theorem:  Let $G:U\times V\rightarrow Y$ be a $C^1$ map where $U \subset X, V\subset Y$ are Banach spaces.  Let $z_0 = G(x_0,y_0)$ and suppose $D^{(2)}G_{(x_0,y_0)}$ is invertible as an element of $L(Y)$.  Then there exist $\delta,\epsilon$ so that for all $x$ with $\| x - x_0 \| < \delta$ there is a unique $y(x)$ with $\| y - y_0 \| < \epsilon$ so that $z_0 = G(x, y(x))$ for all $x$ with $\| x - x_0 \| < \delta$.  Moreover $y \in C^1$ and $D_{x_0} y = | D^{(2)} G_{(x_0,y_0)}|^{-1} D^{(1)} G_{(x_0,y_0)}$.

Pick any covering of $\mathbf{R}^n$ with radius 1 open sets of polydisk type $I \times B^{n-1}(0,1)$ and choose a partition of unity based on this covering.  We can use this covering to reduce to proving the co-area formula for $g$ supported in a ball of radius 1.  So now just assume that support of $g$ is contained in such a polydisk, say $[0,1]\times B^{n-1}(0,1)$.  Now choose a dense countable set $\{ z_i \}$ in the ball with the additional property that $\det|D_{(x_2,\dots,x_n)} \ph(z_i)i| \not = 0$ the implicit function theorem for each of these points to obtain pairs of numbers $(\delta_i,\epsilon_i)$ promised by the theorem.  Use the compact closure of the polydisk to choose a subcover. Now we have a finite set of polydisks with centers $z_1, \dots, z_N$ that cover the support of $g$ with the additional property that the implicit function theorem guarantees that $d/dx \phi(z_j) \not = 0$.  It thus suffices to prove the co-area formula for one of these $N$ open polydisks.

Now it’s a matter of interpreting the implicit function theorem.  Geometrically it is producing a local parametrization of the fibers of the map $late \phi: \mathbf{R}^n\rightarrow \mathbf{R}$.

The implicit function theorem provides us with $(x,y_1(x),\dots,y_{n-1}(x)) \in \mathbf{R}^n$ such that  such that for $t\in\mathbf{R}$, the $\phi^{-1}(t)$ is parametrized by a hypersurface of the polydisk.

We apply this change of coordinates to the integral, and since the fibers are smooth, in this case the Hausdorff measure agrees with the usual hypersurface measure.

For this problem, the application of the implicit function theorem I am happy with but the rest of the argument is too hand-wavy.  So I will come back to it another time.

## Stanford PhD qual 2010 part I, # 2

Consider the spaces L^p[0,1] 1 <= p < infty. For which p is the unit ball B= {f: ||f||_p <= 1} weakly sequentially compact, i.e. for which p does every sequence in B have a weakly convergent subsequence? For each p, prove or disprove weak sequential compactness.

Off the top of my head, Banach-Alaoglu theorem says that the dual space of a Banach space will have the unit ball be weakly compact so every L^p that is the dual space of a Banach space will have this property. The Riesz Representation theorem should give us that the dual space of L^q[0,1] is L^p[0,1] when q>0 and 1/p+1/q=1. So the only exceptional case should be L^1[0,1]. Now it’s time to get some precision to the problem.

Banach-Alaoglu Theorem:  Let $X$ be a Banach space.  Then the unit ball $\{ \ell \in X^*: \| \ell \| \le \}$ is compact in the weak-* topology.

So this answer is right but there is a precise characterization. More precisely, Barry Simon volume I, p. 446 gives us the characterization that the unit ball of a Banach space is weak-* compact if and only if it is reflexive, i.e. if and only if $X=X**$.  Denote the unit ball of a space by a subscript 1.  There is always a continuous injection $i:X\rightarrow X^{**}$ but for reflexive spaces it is a bicontinuous bijection.  Then $i(X_1)=X^{**}_1$ Banach-Alaoglu theorem tells us $X^{**}$ has a weak-* compact unit ball, therefore its inverse is compact.

The converse is slightly more difficult.  If $X_1$ is compact in the weak-* topology then $i(X_1)$ is compact as the continuous image of a compact set hence closed.  We need a theorem that says $i(X_1)$ is dense in $X^{**}_1$ which implies $i(X_1)$ is all of $X^{**}_1$ because $i(X_1)$ is closed as an image of a continuous map.  Then for any nonzero $f \in X^{**}$ we have $f/\| f \| \in X^{**}_1$ so $X$ is reflexive.

Which $L^p[0,1]$ are reflexive is then the issue.  The Riesz Representation Theorem for $L^p[0,1]$ with $1 < p <\infty$ and $1/p+1/q=1$ says that for any $\Gamma \in L^p[0,1]^*$ there exists $g\in L^q[0,1]$ such that $\Gamma(f) = \int fg dx$.  This shows that $(L^p)^* \subset$latex L^q$. We also have $(L^q)^* \subset L^p$ by applying the Riesz Representation Theorem again. Since $L^p \subset (L^p)^{**}$ generally, $L^p \subset (L^p)^{**} \subset L^p$ we conclude $L^p$ is reflexive. It remains then to show that $L^1$ is not reflexive. $(L^1)^* = L^\infty$ but say the delta measure at 1/2 produces a bounded linear functional of $L^\infty$ which is not representable by an integrable function. So the characterization above ensures that $L^1$ is not reflexive. Advertisements ## Stanford math PhD qual 2011 part 1, #5 Let $T = (R/2 pi Z)$ and for $f \in L^2(T^n)$ let $\hat{f}(k) = (2\pi)^{-n/2} \int_{T^n} f(x) e^{-i x k} dx$ be the Fourier coefficients of $f$. For $s>r\ge , m>=0, H^m(T^n)$ is the subset of L^2(T^n) with finite$\sum (1+|k|^2)^m |\hat f(k)|^2 < infty$and norm $\| f \|_m = \sum ( 1 + |k|^2 )^m | \hat{f}(k) |^2$. $P: H^s(T^n) -> H^m( T^n)$ is a continuous linear map satisfying $\| u \|_s <= C( \|Pu\|_m + \| u \|_r )$ — call this (1) Show that the nullspace of $P$ is finite dimensional subspace of$H^m$and its range is closed. Proof: If $P:H^s -> H^m$ is a bounded operator satisfying $\| u \|_{r+t} <= C( \| Pu \|_m + \| u \|_r )$ then the nullspace is a closed subspace since it’s the preimage of a closed set. All elements of the nullspace satisfy $\| u \|_{r+t} \le C\|u\|_r$. Now the inclusion map$latexj:H^{r+t}->H^r$is continuous which gives $\| u \|_r <= C_1 \| u \|_{r+t}$. What we have to show is that if $\{ u_j \}$ is a bounded sequence in $Ker(P)$ there must be a convergent subsequence. We can evade the issue at this point and apply a hammer — the Rellich Compactness Theorem says that the inclusion $j:H^{r+t}->H^r$ is actually compact, Our bound $\| u \|_{r+t} \le C \| u \|_r$ tells us that all elements of $Ker(P)$ must lie in $H^{r+t}$. If $\| u_j \|_r \le 1$, then $\| u_j \|_{r+t} <= 1$ and the Rellich compactness theorem then tells us that there is a convergent subsequence in the norm of $H_r$. Therefore the unit ball of $Ker(P)$ is compact, which implies that $Ker(P)$ is finite dimensional. The problem with this solution, of course, is that we applied the Rellich Compactness Theorem and there might be a way to prove the compactness directly. For the second issue, that $Ran(P)$ is closed, I don’t have a clear idea yet. Let $y$ be an element of the closure of the range and let $\{Pu_j \}$ be a Cauchy sequence that converges to it so for $\epsilon > 0$ assume $\| Pu_{n+1} - P u_n \|_m < 2^{-n} \epsilon$. Then we get $|| u_{n+1} - u_n ||_{r+t} <= C( 2^{-n} epsilon + || u_{n+1} - u_n ||_r)$ Continuity of the inclusion map $j: H_{r+t} -> H_t$ gives then $\| u_{n+1} - u_n\|_{r+t} <= C ( 2^{-n} epsilon + C_1\| u_{n+1} - u_n \|_{r+t} )$ So one idea is that$C_1 < 1$would allow us to telescope the right side: $\| u_{n+1} - u_n ||_{r+t} <= C ( 2^{-n} epsilon + C_1 C ( 2^{-n} epsilon + C_1 ||u_{n+1} - u_n||_{r+t} )$ then $\| u_{n+1} - u_n \|_{r+t} <= C( 1 + C_1 C + (C_1 C)^2 + ...) 2^{-n} \epsilon$ which then would imply that $\| u_{n+1} - u_n \|_{r} \le \| u_{n+1} - u_n \|_{r+t} <= C_2 2^{-n} \epsilon$ and this is enough to show $\{ u_j \}$ is Cauchy in $H_r$ so there is a limit $u_\infty$. Then continuity of $P$ implies that $y = lim Pu_j = P( lim u_j)$ which shows that $y$ is actually in $Ran(P)$. This is what I had in the first sitting more or less without looking around for ideas. Then I found the following general result whose proof can be extracted from my work above: a bounded injective $T: X\rightarrow Y$ has closed range if and only if there is a constant $c>0$ such that $\|Tx\| \ge c \| x \|$. In one direction, the lower bound condition allows us to show that if $Tx_j$ is Cauchy then $x_j$ is Cauchy by inequality$\| Tx_j – Tx_k \| \ge c \| x_j – x_k \|\$.  This is what I was attempting to capture in my proof above by the telescoping argument.  In the other direction, one applies the Open Mapping Theorem which says that a bijective bounded linear operator has a bounded linear inverse.  The injective $T$ maps onto the range.  The range is closed so it’s complete, and so there exists a map $T^{-1}: Ran(T)\rightarrow X$ with $\| T^{-1} y \| \le c \| y \|$.  Plug in $y=Tx$ to obtain $\| x \| \le c \| Tx\|$.

For the second part of the problem, I can apply this general result to $P | N(P)^{\perp}$ to obtain an injective bounded linear operator above and then apply the result to conclude that $Ran(P)$ is closed.  In any case, the remaining issue is whether my iteration scheme must produce $\| P u \| \ge c\| u \|$.

## An elementary argument for sup bounds from a differential inequality?

The question arises from trying to understand this Donaldson-Sun paper.  The issue is given a differential inequality $\Delta_{\bar{\partial}} |s| \le |s|$ for the sections of some line bundle and the Sobolev inequality $\| f \|_{2+\epsilon} \le C_1 \| \nabla f \|_2 + C_2 \| f \|_2$ to conclude that $\| s \|_\infty \le C \| s \|_2$ and $\| \nabla s \|_{\infty} \le C \| s \|_2$ (the constants are not the same).  There is the Nash-Moser iteration scheme that could be applied and is what is used by workers in the field.  On the other hand, one wonders if there is an elementary argument that could work.  Of course the likelihood is probably small because otherwise it would have been adopted, but here is an approach.  Start with $\langle \| \nabla s \|_2 \le \Delta |s|, |s| \rangle \le \|s\|_2^2$ and plug this into the Sobolev inequality collecting terms.

$\| \nabla s \|_{2+\epsilon} \le C \| s \|_2$

Well I was hoping that some elementary argument might give us $L^{\infty}$ bounds but this is too naive.  So this is a situation that has been explicitly addressed by Karen Uhlenbeck on her great paper on removable singularities of Yang-Mills fields, which gives me an opportunity to revisit this issue.  She quotes the following Theorem from Morrey’s Multiple Integrals in the Calculus of Variations:

The exact case she addresses is: weak inequality $\Delta f \le b f$ for $b \in L^q(B(x_0,a_0))$ for $q>n/2$ and $f \ge 0$ and $f^{\gamma} \in L^2_1(B(x_0,a_0))$ for $0\le \gamma \le 1$.  The conclusion of her Theorem 3.2 is that $f$ is bounded in the domains interior to $B(x_0,a_0)$ and the constants $K_1$ depends uniformly on $n,q,\gamma$ and $a_0^{q-n/2} \int b^q$.

$f^{\gamma}(x)|^2 \le K a_0^{-r} \int | f^{\gamma}(y)|^2$

First she proves in a Lemma which is also useful in this case.

$\int d(uf^p) le \int |p-1| p (2p-1) \Delta u^2| + (du)^2] f^{2p} + p^2/2p - (\int b^{2/n})^{n/2}(\int (u f^p)^{\nu})^{2/\nu}$

## Peter Li’s Sobolev 1-norm to Sobolev 2-norm lemma

This note is actually about a real analysis exercise because I could not understand how Peter Li produced the zero integral condition in this paper.  The setting is a compact riemannian manifold $M$ with riemannian volume and the issue is Sobolev inequality $C_2 \| f \|_{2m/(m-2)} \le \| \nabla f \|_2^2$ from $C_0 \inf_{a\in \mathbf{R}} \| f - a \|_{\frac{m}{m-1}}^m \le \| \nabla f \|_1^m$.  The issue is the following step.

If $g$ satisfies $\int_M (sign g) |g|^\frac{1}{m-1} =0$ then $\| g \|_{\frac{m}{m-1}} \le \inf_{a\in\mathbf{R}} \| g - a \|$.  I just realized that the condition is just the critical point condition for the function $F(a) = \int |g-a|^p$.  The derivative of the function $x\rightarrow |x|^p=(x^2)^{p/2}$ is $x \rightarrow p x |x|^{p-2} = (sign x) |x|^{p-1}$ so the condition $F'(a) = 0$ is exactly $\int sign ( g-a) |g-a|^{p-1} = 0$.  In his Lemma Peter Li is explicitly setting $a=0$ in the condition.  It’s elementary and all but it was not obvious to me until I did this exercise.

Just to follow Peter Li’s argument, we want to compute $C \| f \|_{2m/(m-2)}^{2m(m-1)/(m-2)} = C_0 [ \int |f|^{2m/m-2} ]^{(m-2)/2m \cdot 2m (m-1)/(m-2)} = C_0 [ \int |g|^{2m/(m-2) \cdot (m-2)/2(m-1) ]^{m-1} = C_0 [ \int |g|^{2m/(m-2) \cdot (m-2)/2(m-1) ]^{m-1} = C_0 [ \int |g|^{m/(m-1)} ]^{(m-1)/m \dot m} = C_0 \| g \|_{m/(m-1)}^m$.  Then use the Sobolev 1-norm inequality on $g$.

$C_0 \| g \|_{m/(m-1)}^m \le \| \nabla g \|_1^m$.

Compute the gradient of $g$ directly
$\nabla g = \frac{2(m-1)}{m-2} |f|^{m/(m-2)} \nabla f$
Then plug this into the Sobolev 1-norm inequality, apply Schwarz inequality on the

$|f|^{m/(m-2)}$

term and rewrite that as an $L^p$ norm of $f$, divide both sides by $\| f \|_{2m/(m-2)}^{2m(m-1)/(m-2)}$ and get the Sobolev inequality for 2-norm of Peter Li’s lemma.

For general $f \in L^2_$ which do not satisfy the condition $\int (sign f)|f|^{2/(m-1)} = 0$ the idea is to choose $k\in\mathbf{R}$ such that $\int sign (f-k) | f - k |^{2/(m-1)} =0$ and apply Peter Li’s Lemma 1:

$\| f-k \|_{2m/(m-2)}^2 \le \| \nabla f \|_2^2$

Apply the convexity bound $(a/2+b/2)^p \le a^p + b^p$ and some other elementary estimates to get the general case.  Anyway Peter Li’s paper has many results on bounds on eigenvalues on differential forms using the inequality.  I just wanted to get some idea of what is involved in proving the Sobolev 2-norm inequality from the 1-norm version.

So my intention was to have a better understanding of the Sobolev inequality itself.  This lemma of Peter Li is telling us how to go from the stronger 1-norm Sobolev inequality to the 2-norm.  I found some insight about this issue from the paper of Talenti who proved the best constant for the Sobolev inequality on Euclidean spaces.  The connection can be understood by setting $f=1_E$ where $E$ is some bounded domain.  The Sobolev inequality then becomes $Vol(E)^{(m-1)/m} \le C Area(\partial E)$.  That’s the insight I really needed for the moment.