Archive for the ‘Uncategorized’ Category

This is totally standard material that I am reviewing because I have forgotten much of the mathematics that I had learned 20 years ago.  These are from notes by Alex Massarenti that I found here.


We care about projective algebraic varieties.  We know that an analytic subvariety of the complex projective space is algebraic by the theorem of Chow.  We want to know if a complex Kahler manifold is going to be algebraic.  A an even dimensional riemannian manifold (M,g) is a Kahler manifold if there is a real endomorphism J of the tangent bundle such that J^2=-1, i.e. an almost complex structure satisfying g(Ju,Jv)=g(u,v) and the form \omega(u,v) = g(Ju,v), is closed d\omega=0.

Let M be a compact complex manifold and L a holomorphic line bundle on it.  Any subspace W of H^0(M,\mathcal{O}(L)) determines a linear system |W| = \{ s \}_{w\in W} of divisors on latex M$ by their zero sets.  Since M is compact, (s) = (s') if and only if s=\lambda s' for some \lambda \in \mathbf{C}, and |W| is parametrized by \mathbf{P}(W).  If the linear system |W| has no base point for any point p the set of sections s vanishing at p forms a hyperplane H_p \in \mathbf{P}(W) and we have a map \phi: M \rightarrow \mathbf{P}(W)^*.  More explicitly let \{ s_0, \dots, s_N \} form a basis of W and U \subset M open and \psi a local trivialization of L on M,

\p(p) = [ s_0(p) : \cdots : s_N(p) ]

Clearly \psi is holomorphic and L=\psi^*|W(\mathcal{O}_{\mathbf{P}^N}(1) ).

The above gives a sketchy description of the way in which one embeds the complex manifold into complex projective space with sections of a line bundle.  I will have to come back to work out the details so this is clearer.  But I want to get to the deep mathematics issue here, which is Kodaira’s great 1954 embedding theorem.

A line bundle L on M is defined to be positive if the first Chern class c_1(L)>0 in the sense that its representative in H^2_{dR}(M) is positive.

Let M be a compact complex manifold and L a positive line bundle on M.  There exists \bar{k} such that for k\ge \bar{k} the map \phi_{jk} : M \rightarrow \mathbf{P}(H^0(M,L^k)^*) is an embedding.



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Let \pi be a unitary representation of U(n).

In the last episode, we have gone through the argument that if \Phi is a character that satisfies (i) it is a finite Fourier series with integer coefficients, (ii) it is skew symmetric under the action of the permutation group W, and (iii) for \pi,\pi different irreducible representations then C \int \Phi_{\pi} \Phi_{\pi'} dx = \delta_{\pi \pi'}.  We showed that the tilde-map construction \sim: \hat{D} \rightarrow \hat{D} given by \tilde{\chi} = \sum_{w\in W} \epsilon(w) \chi^w then in fact any \Phi satisfying the conditions (i) and (ii) and (iii) must be of the form \pm \chi_m for m=(m_1,\dots,m_n) with decreasing values.  Now the problem is to show that every m that satisfies there conditions occur for \Phi satisfying (i), (ii) and (iii) as we vary the representation \pi.

If q =(q_1,\dots,q_n) does not occur, the idea is to produce a contradiction with the completeness of L^2(G)^{inv}.  Consider \Theta_D = \chi_q/\Delta which is invariant under W, belongs to L^2(G)^{inv} and is orthogonal to (\Theta,\Theta_{\pi})_G = 0 for as \pi varies over irreducible representations and this will contradict the completeness of \Theta in L^2(G)^{inv}.

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Our story continues with having reached the main technical idea according to Varadarajan of Weyl’s Character Formula, which is that one should consider the orthogonality relation between characters of irreducible representations \int \theta_g \theta_{g'} dx = \delta_{g g'} and decompose G=U(n) along two different directions.  On one hand, along the diagonal elements D and on the other hand along the conjugacy classes each of which is going to be a submanifold of some sort in G.


The integration formula that Weyl uses for the orthogonality relation is valid for points that are ‘nonsingular’ in a sense.  The condition is that the conjugacy class must have maximum possible dimension.  At a point x\in G we want to understand the condition on the conjugacy class to which x belongs, recall that x,y belong to the conjugacy class if and only if there exists g\in G such that x= g y g^{-1}.  We next want to characterize the conjugacy class as exactly the coset space G/G_x where G_x is the centralizer of x, i.e. those g\in G that commute with x.  Let’s be elementary about this, so if g \in G_x the the condition x = g y g^{-1} is the same as xg = gx = gy which occurs if and only if x=y. Therefore we can exactly identify the distinct elements of the conjugacy class of x with G/G_x.  The two points of note is first that y \rightarrow y x y^{-1} maps y on to the conjugacy class of x, and a different argument for distinctness of points of the conjugacy class is y_1 x y_1^{-1} = y_2 x y_2^{-1} implies y_2^{-1} y_1 x = x y_2^{-1} y_1 which implies that $latex  y_2^{-1} y_1 \in G_x$ and y1 \in y_2 G_x.

Now let us try to understand the dimension of this space G/G_x.  If \lambda_1, \dots, \lambda_r are distinct eigenvalues of x with multiplicities m_1, \dots, m_r then we want to calculate the dimension of G_x in terms of these multiplicities.  Elements of G_x preserve the eigenspaces for x v = \lambda v implies both $latex gx v = \lambda gv as well as x gv = g \lambda v = \lambda gv.  Therefore for a fixed \lambda_i we certainly have that at most m_i^2 components contribute to dim(G_x).  Therefore m_1^2 + m_2^2 +\dots + m_r^2 \ge dim(G_x). On the other hand, Varadarajan claims that this is an equality, i.e. that m_1^2 + \dots + m_r^2 = dim(G_x).  In other words, that the dimension of matrices that commute with the matrix x of the multiplicity m_j eigenspace of eigenvalue \lambda_j has dimension m_j^2.  This is not clear to me yet why.  This dimension, m_j^2 seems too large at first glance but I am quite confident that my worries are unnecessary so skipping the issue:

m_1^2 + \dots + m_r^2 = dim(G_x)

Our original question is when dim(G_x) is maximized.  We have the constraint that m_1 + \dots + m_r = n.   Therefore m_1^2 + \dots + m_r^2 \ge n.

Now m_1^2 + \dots + m_r^2 \ge n and the two sides are equal when r=n and m_1 = \dots = m_n = 1.  Actually this last part tells us that we needed only m_1^2 + \dots + m_r^2 \ge dim(G_x) and not equality.

Since dim(G/G_x) is maximum if and only if dim(G_x) is minimum, above can be used to conclude dim(G/G_x) is maximum if and only if x has n distinct eigenvalues, which is the same as the characteristic polynomial having n distinct roots.


The characteristic polynomial with a variable det(T -y) = T^n - c_1(y) T^{n-1} + c_2(y) T^{n-2} + \dots \pm c_n(y) = \prod (T-\lambda_j).  Now \prod_{i \not= j} (\lambda_i - \lambda_j) is invariant under permutations and by Newton’s theorem is a function of the symmetric elementary functions P(\sigma_1, \dots, \sigma_n) where for example sigma_1(X_1, \dots, X_n) = \sum X_i and so on.  The discriminant \delta(y) = P(c_1(y), \dots, c_n(y)) is such that \delta(y) \not=0 if and only if latex x$ has n distinct roots.  We call x\in G is regular if x has n distinct eigenvalues and denote the set of regular values as G' and call the complement G-G' the singular set.  The key point is that the singular set is small.  More precisely, the Haar measure which is used for all the integration is absolutely continuous with respect to the Lebesgue measure in coordinates and since the singular set is defined as the zero set of an analytic function, it has Haar measure zero.


Recall that the Character Formula is that the irreducible characters of U(n) are in natural one-to-one correspondence with m= (m_1, \dots, m_n) where m_1 > m_2 > \dots > m_r.  The character

\Theta_m = (\sum_{w \in W} \epsilon(w) \chi^w_m)/Delta

where \chi_m is the character t = diag (t_1,\dots,t_r) \rightarrow t^m = t_1^{m_1} \dots t_r^{m_r}


\Delta(t) = \prod_{i \not= j} (t_i - t_j)

For an irreducible representation \pi define ( \Theta_n)_D \Delta = \Phi_{\pi}.  Then the following are relatively easily established:

(i) \Phi_{\pi} has finite Fourier series with integer coefficients

(ii) \Phi_{\pi}^w = \epsilon(w) \Phi_{\pi}

(iii) \frac{1}{n!} \int \Phi_{\pi} \Phi_{\pi'} dt = \delta_{[\pi][\pi']

Since \Delta = \prod_{i\not=j}(t_i - t_j) a finite Fourier series with integer coefficients which is skew symmetric with integer coefficients from the formula while \Theta_\pi is a symmetric finite Fourier series with integer coefficients (i) and (ii) are clear while (iii) is just the orthogonality relation.

The interesting and quite sophisticated part of the proof is to construct all possible \Phi that satisfy the three conditions.

I am going to come back to the pieces maybe tomorrow.

Ok now it is tomorrow, March 18 2018.

For any diagonal character \chi \in \hat{D} the alternating sum

\tilde{\chi} = \sum_{w \in W} \epsilon(w) \chi^w

is a finite Fourier sum and  if s \in W then \tilde{\chi}^s = \sum_{w \in W} \epsilon(w) \chi^{w s} = \sum_{w \in W} \epsilon(w) \epsilon( s )^2 \chi^{ ws } = \sum_{w \in W} \epsilon(s) \epsilon( ws }= \epsilon(s) \sum_{w} \epsilon(ws) \chi^{ws} = \epsilon(s) \sum_{w} \epsilon(w) \chi^w.  For the first part, we just inserted 1 = \epsilon(s)^2 and then we changed the index of the sum over all the permutations in W which shows (ii).

Any character satisfying (i) and (ii) must be of the form

\Phi = \sum_{m_1 > \dots > m_n} c(m_1, \dots, m_n) \tilde{\chi}_{m_1, \dots, m_n}

where the coefficients are integers and

\| \Phi \|^2 = \sum |c(m_1, \dots, m_n)|^2



If \chi_1 and \chi_2 are in different W-orbits then ( \chi_1 | \chi_2 ) = 0.  Recall that the characters are elements of \hat{D} and D = T^n the n-torus.  If \chi_1 and \chi_2 are in different W-orbits then there are no permutations that can equate \chi_1^w = \chi_2.  The orthogonality is actually the outcome of a separate result known as ‘the orthogonality relation for characters’: (\theta_a|\theta_b) = \delta_{ab}.  If the \chi_1 = \chi_2^{w0} then \tilde{\chi_2} = \epsilon(w_0) \chi_1.


Let pi be an irreducible representation of U(n).
So the key issue in the Weyl Character Formula is to consider the three conditions (i) Phi is a finite Fourier series with integer coefficients, (ii) Phi_pi^w = sign(w) Phi_pi, (iii) C \int Phi_pi Phi_{pi’} = delta_{pi pi’}.
For any character chi whatsoever, we can construct a character
tilde{chi} = sum_{w in W} sign(w) chi^w
and this will satisfy (i) obviously and (ii) by taking for any permutation s and tilde{chi}^s = sum_w sign(s)^2 sign(w) chi^{sw} = sign(s) sum_w sign(sw) chi^{sw} = sign(s) sum_{sw} sign(sw) chi^{sw}. The general orthogonality relation for characters guarantees that (tilde{chi}_1}|\tilde{chi}_2) = 0 when chi_1 and chi_2 are in different W-orbits. If they are in the same orbit, chi_2 = chi_1^w then
tilde{chi_2} = tilde{ chi_1^ w } = sign(w) chi_1 by (ii)
We have D the n-torus in the diagonal of U(n) and the characters in the dual \hat{D}. So far we have produced a map \tilde: \hat{D} -> { characters that satisfy (i), (ii) } and we analyzed what happens in the image of this map under the action of permutations W.
Ok, now we step back and recall that all the characters of U(n) can be written as diag(t_1,…,t_n)->t^m = t_1^{m_1} … t_n^{m_n} and we consider what this tilde-map is doing in terms of these.
The tilde-map will zero out those chi_m which have repeated values in m=(m_1,…,m_n) because if we take m_i=m_j then we can find permutation w swapping i and j indices leaving all else the same which has sign -1 so chi_m^w = sign(w) chi_m = – chi_m so chi_m=0. This imples the image of the tilde-map only contains combinations of chi_m with distinct elements. Then we can find a permutation that re-orders the elements of (m_1,…,m_n) in decreasing order.
So ANY Phi satisfying (i) and (ii) a combination with integer coefficients and finite sum
sum c(m) \tilde{ chi_m}
and furthermore || Phi ||^2 = 1. This last condition is going to force only one of the coefficients to be nonzero because the coefficients are all integers. and that coefficient will be +/-1.
So far we have shown any Phi satisfying (i) and (ii) will pick out an m with a coefficient +/- 1. We still have to show that all possible m occur as the image of the tilde map with Phi satisfying (i) and (ii).



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Hermann Weyl is one of the greatest mathematicians of the early twentieth century and since having wasted a couple of decades lost in fog not recognizing what is really important and what is not, thereby not realizing that not even physics is as important as mathematics, I want to regain my mathematical footing.  So I thought that while going through Lie group theory, I learn something about Weyl’s work on representations of unitary groups.

Recall that U(n) is the set of n\times n matrices A with complex entries with A^* A= 1.  Here A^* is the conjugate transpose.  We denote by D the diagonal matrices of U(n)  Recall that a representation T of a Lie group G is a group homomorphism T: G \rightarrow End(V) = Hom(V,V).  Recall that a character of a representation T is a map \theta: G \rightarrow \mathbf{C} defined to be the trace of the matrix T(g), so \theta(g) = trace (T(g))

So here is the theorem I want to understand, the Weyl Character Formula for the group.  The work of Weyl is deep and extensive, but I want to go slow so I can understand things clearly.

The irreducible characters of U(n) are in natural one-to-one correspondence with n-tuples (m_1,\dots,m_n) with $m_1 > \dots > m_n$.  The character \Theta^{(m_1,\dots, m_n)} is given on the diagonal matrices by

\Theta^{(m_1,\dots,m_n)}|_D = \sum_{w\in W}  \epsilon(w) \chi^{w}_{(m_1,\dots,m_n)})/\Delta

where W is the set of permutation of the indices, \chi_{m_1,\dots,m_n} is the character t=diag(t_1,\dots,t_n) \rightarrow t_1^{m_1} \dots t_n^{m_n}, and

\Delta(t) = \prod_{1 \le i < j \le n} (t_i - t_j)

I am following Varadarajan’s An Introduction to Harmonic Analysis on Semi-Simple Lie Groups and follow  his presentation.

The permutation group W acts naturally on the diagonal matrices D as well as the dual group \tilde{D} of characters of D, homomorphisms of D \rightarrow S^1, so (w T)( diag(t_1,\dots,t_n)) = T ( diag(t_{w(1)}, \dots, t_{w_n})).

A conjugacy class \gamma is G/E where the equivalence relation on G is defined by a \sim b if and only if there exists g\in G such that b = g b g^{-1} and since unitary matrices are diagonalizable, every conjugacy class meets the diagonal D, and \gamma \cap W is a single W-orbit, namely the orbit of permutations of a fixed set of eigenvalues \lambda_1, \dots, \lambda_n.

For any class function f on G, i.e. a function that is constant on each conjugacy class, and denote by f_D, obviously f and f_D are in one-to-one correspondence.  Furthermore, f_D must be invariant under the action of permutations W.

If \pi is a finite dimensional unitary representation of G the restriction of \pi to D is a direct sum of characters of D.

Let’s unpack the above a little bit more.  The Peter-Weyl theorem asserts that the irreducible unitary representations of a compact Lie group are all finite dimensional, the irreducible characters form an orthonormal basis of L^2(G)^{inv} composed of L^2 functions on G and a subspace L \subset V is \pi-invariant if $\pi(g) L \subset L$ for all g\in G.  The Peter-Weyl theorem also tells us that L^2(G) decomposes as a direct sum of irreducible representations. Although this is using too big a hammer, the Peter-Weyl Theorem has the corollary that the restriction of \pi to the diagonal matrices of U(n) is a direct sum of characters

(\theta_n)_D = \sum_j m_j \chi_j

with integer coefficients m_j \ge 0  The diagonal elements of a unitary matrix are of form (e^{it_1}, \dots, e^{it_n}), to put things in context.  So the irreducible representations on the diagonal of the group are finite Fourier series with non-negative integral coefficients, and it is moreover symmetric or W-invariant.

The idea behind Weyl’s method is to start with the orthogonality relations \int \theta_g \theta_{g'} dx = \delta_{g g'} for g,g' \in G.  The integral is over the entirety of G but Weyl breaks up the integral into diagonal D and then the orbits of the conjugacy as a double integral.

So this is the crux of the matter — if we can decompose the integral in the above manner, the character formula will emerge.

Now I am tired already and will return to the rest of the proof later.


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Let T be the 2-torus \{ diag(e^{i x}, e^{iy}: x,y \in [0,2 \pi]\} and let S=\{ diag( e^{it}, e^{it\sqrt{2}} ): t\in \mathbf{R} \}.  This is a subgroup because e^{i(t+s)} = e^{it} e^{is}.  The problem is to prove that the closure of S is not 1-dimensional.  Deduce that S is dense in T.

This problem seems equivalent to Weyl’s equidistribution theorem: the points \alpha k mod 1 fill up the interval [0,1] if and only if \alpha is irrational.  This is the idea.  Fix x \in [0,1] and we want to approximate any given y\in [0,1] by k x \sqrt{2} mod 1 for some integer k.  Ok actually this is a different proof of the Weyl equidistribution.

The map \phi:\mathbf{R}\rightarrow \phi(\mathbf{R}) is obviously surjective.  Now diag( e^{it},e^{it\sqrt{2}}) = diag( e^{it},e^{is\sqrt{2}}) implies two things: that s\sqrt{2}=t\sqrt{2} \mod 2\pi and s = t mod 2\pi.  The first equation implies there exists an integer k such that s\sqrt{2} = t\sqrt{2} + 2\pi k and therefore s=t + 2\pi k/\sqrt{2} then (s-t)/2\pi = k/\sqrt{2} which cannot be an integer unless s-t=0 so the second equation cannot hold.  Thus \phi is injective.  Now we can apply the result that a bijective homomorphism of locally compact Hausdorff topological groups must be a homeomorphism to conclude that \phi(\mathbf{R}) is a homeomorphism.  Therefore \phi(\mathbf{R}) cannot be closed, for otherwise it must be compact, which is impossible since \mathbf{R} is non-compact.  The closure \bar{\phi(\mathbf{R})} is a compact subgroup of T.  Assume this closure is one-dimensional.  Then it must be realized by the exponential map from a one-dimensional subalgebra of the Lie algebra of the torus by the exponential map, so there exists a,b \in \mathbf{R} such that \bar{\phi(\mathbf{R})} = \{ diag( e^{iat}, e^{ibt}), t\in\mathbf{R} \}.  This is only possible by the previous argument when a/b \in \mathbf{Q}.  Therefore \bar{S} cannot be one-dimensional.  Therefore it must be two dimensional.  But the only 2-dimensional subalgebra of the commutative Lie algebra \mathbf{R}^2 is the entire space.  This implies \bar{S} = T and S is dense in T.


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An exercise from Knapp’s ‘Beyond First’ book on Lie groups.  This is completely my own.  I did get a hint that the Pettis Theorem would be needed.  The Pettis Theorem is that for a non-meager subset Uof a Baire topological group U^{-1} U is a neighborhood of the identity.

Let \pi: G_1 \rightarrow G_2 be a bijective homomorphism between locally compact topological groups.  Let \sigma = \pi^{-1} which we claim is also a homomorphism: \sigma(v_1 v_2) = \sigma( \pi(u_1) \pi(u_2) ) = \sigma( \pi( u_1 u_2 ) ) = u_1 u_2 = \sigma( v_1) \sigma(v_2) so \sigma is a homomorphism.

Let U \subset G_1 be open which assures us U U^{-1} is open.  We would like to show \pi(U) is open.  Regardless of whether it is open or closed, it must be non-meager.  Then \pi(U) \pi(U)^{-1} is an open neighborhood of the identity regardless of whether \pi(U) is a priori open.  Then \pi(U U^{-1}) = \pi(U) \pi(U)^{-1} is open since \pi is continuous.  Now U^{-1} is open.  We want to conclude from \pi(U)\pi(U^{-1}) open to \pi(U) open.   Suppose \pi(U) is not open.  Then neither is \pi(U^{-1}) and hence \pi(U)\pi(U^{-1}) is not open.


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Lie groups

These are notes from Cheeger-Ebin.  A Lie group G is a manifold for which (x,y) \rightarrow xy^{-1} is smooth and a Lie algebra is a vector space with a pairing [,] satisfying linearity, anticommutativity and the Jacobi identity [ A, [B,C]] + [B,[C,A]] + [C,[A,B]]=0. Vector fields on manifolds form an infinite dimensional Lie group with commutators satisfying the above axioms.  The tangent space of a Lie group at a point is isomorphic to the left invariant vector fields on it and therefore has the structure of a Lie algebra.  Any finite dimensional Lie algebra can be shown to be the Lie algebra of some Lie group.  A subalgebra of a Lie algebra considered as the left invariant vector fields on a Lie group defines an involutive (meaning the brackets stay within the subspace) distribution so the Frobenius theorem guarantees a submanifold and the maximal submanifold through the identity is a subgroup.  Conversely a subgroup of a Lie group then the tangent space is a Lie subalgebra which is a normal subalgebra if and only if it is an ideal.  Since any one dimensional subalgebra is involutive, the corresponding subgroup is a one-parameter subgroup.  A compact G admits a bi-invariant metric.

If \langle , \rangle is a left-invariant metric:
(a) \nabla_X Y = \frac{1}{2}( [X,Y] - (ad_{X})_* Y - (ad_{Y})_* X )

(b) \langle R(X,Y) Z, W \rangle = \langle \nabla_X Z, \nabla_W \rangle - \langle \nabla_Y Z, \nabla_X W \rangle - \langle \nabla_{[X,Y]} Z, W \rangle

(c) \langle R(X,Y) Y, X \rangle = \| [ (ad_X)^* Y - (ad_Y)^* X ] \|^2 - ((ad_X)^*(X), (ad_Y)^*(Y)) - \frac{3}{4} \| [X,Y] \|^2 - \frac{1}{2}([[X,Y],Y],X) -\frac{1}{2}([[Y,X],X],Y)

Ok, so the homogeneous spaces are those which occur as G/H for some subgroup H and everyone’s favourite homogeneous manifolds are spheres and hyperbolic spaces which have constant curvatures.  The map \pi: G \rightarrow G/H is a riemannian submersion defined by the property that d\pi|_H is an isometry.

An interesting result is that if \pi: M\rightarrow N is a riemannian submersion and \gamma: [0,1]\rightarrow N and \bar{\gamma}:[0,1]\rightarrow M is a horizontal lift then \gamma is a geodesic if and only if \bar{\gamma} is.

The most interesting homogeneous space is probably the upper half complex plane with  SL(2,\mathbf{R}) consisting of matrices with determinant 1 by Moebius transformations z\rightarrow (az+b)/(cz+d).  Grassmannians Gr(n,r) = O(n)/(O(n-r)\times O(r)), projective spaces are homogeneous, etc.  The entire Thurston Geometrization program is to decompose compact three dimensional manifolds into homogeneous pieces.  Yau likes to emphasize the search for special metrics like Einstein metrics Ric = c gMost known Einstein metrics with c>0 are homogeneous.


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