The tangent space and differentials

 Preliminary exam prep

Let $M,N$ be smooth $n$-manifolds. Here we discuss different definitions of the tangent space and differentials, or pushforwards, of smooth maps $f:M\to N$.

Derivations (Lee)

Definition: A derivation of $M$ at $p\in M$ is a linear map $v:C^\infty(M)\to \R$ such that for all $f,g\in C^\infty(M)$,
$$v(fg) = f(p)v(g) + g(p)v(f).$$
The tangent space $T_pM$ to $M$ at $p$ is the set of all derivations of $M$ at $p$.

Given a smooth map $F:M\to N$ and $p\in M$, define the differential $dF_p:T_pM\to T_{f(p)}N$, which, for $v\in T_pM$ and $f\in C^\infty(N)$ acts as
$$dF_p(v)(f) = v(f\circ F)\in \R.$$

Dual of cotangent (Hitchin)

Definition: Let $Z_p\subset C^\infty(M)$ be the functions whose derivative vanishes at $p\in M$. The cotangent space $T_p^*M$ to $M$ at $P$ is the quotient space $C^\infty(M)/Z_p$. The tangent space to $M$ at $P$ is the dual of the cotangent space $T_pM = (T_p^*M)^* = \Hom(T_p^*M,\R)$.

Given a smooth map $F:M\to N$ and $p\in M$, define the differential
$$\begin{array}{r c l} dF_p\ :\ T_pM & \to & T_{F(p)}N, \\ \left(f:C^\infty(M)/Z_p \to \R\right) & \mapsto & \left(\begin{array}{r c l} g\ :\ C^\infty(N)/Z_{F(p)} & \to & \R, \\ h & \mapsto & f(h\circ F). \end{array}\right) \end{array}$$
This definition makes clear the relation to the first approach. Since $h\not\in Z_{F(p)}$, the derivative of $h$ does not vanish at $F(p)$. Hence the derivative of $h\circ F$ at $p$, which is the derivative of $h$ at $F(p)$ multiplied by the derivative of $F$ at $p$, does not a priori vanish at $p$.

Derivative of chart map (Guillemin and Pollack)

Definition: Let $f:\R^n\to \R^m$ be a smooth map. Then the derivative of $f$ at $x\in \R^n$ in the direction $y\in \R^n$ is defined as
$$df_x(y) = \lim_{h\to 0}\left[\frac{f(x+yh)-f(x)}h\right].$$
Given $x\in M$ and charts $\varphi:\R^n\to M\subset \R^m$, the tangent space to $M$ at $p$ is the image $T_pM = d\varphi_0(\R^n)$, where we assume $\varphi(0)=p$.

Given a smooth map $F:M\to N$ and charts $\varphi:\R^n\to M$, $\psi:\R^n\to N$, with $\varphi(0)=p$ and $\psi(0)=F(p)$, define the differential $dF_p:T_pM\to T_{F(p)}N$ via the diagrams below.

Here $h = \psi^{-1}\circ F\circ \varphi$, so $dh_0$ is well-defined. Hence $dF_p = d\psi_0\circ dh_0\circ d\varphi_0^{-1}$ is also well-defined.

Sometimes the differential is referred to as the pushforward, in which case it is denoted by $(F_*)_p$.

References: Lee (Introduction to Smooth Manifolds, Chapter 3), Hitchin (Differentiable manifolds, Chapter 3.2), Guillemin and Pollack (Differential topology, Chapter 1.2)

Degree and orientation

 Preliminary exam prep

Topology

 Recall that a topological manifold is a Hausdroff space in which every point has a neighborhood homeomorphic to $\R^n$ for some $n$. An orientation on $M$ is a choice of basis of $\R^n$ in each neighborhood such that every path in $M$ keeps the same orientation in each neighborhood. Every manifold $M\owns x$ appears in a long exact sequence (via relative homology) with three terms
$$H_n(M-\{x\}) \tov{f} H_n(M) \tov{g} H_n(M,M-\{x\}).$$
The first term is 0, because removing a point from an $n$-dimensional space leaves only its $(n-1)$-skeleton, which is at most $(n-1)$-dimensional. For $U$ a neighborhood of $x$ in $M$, the last term (via excision) is
$$H_n(M-U^c, M-\{x\}-U^c) = H_n(U, U-\{x\})\cong H_n(\R^n, \R^n-\{x\}) \cong H_n(\R^n, S^{n-1}),$$
which in turn fits into a long exact sequence whose interesting part is
$$H_n(\R^n)\to H_n(\R^n,S^{n-1}) \to H_{n-1}(S^{n-1})\to H_{n-1}(\R^n),$$
and since the first and last terms are zero, $H_n(M,M-\{x\})=\Z$. Since $f$ is zero, $g$ into $\Z$ must be injective, meaning that $H_n(M)=\Z$ or 0.

Theorem: Let $M$ be a connected compact (without boundary) $n$-manifold. Then

1. if $M$ is orientable, $g$ is an isomorphism for all $x\in M$, and
2. if $M$ is not orientable, $g=0$.

Definition: Let $f:M\to N$ be a map of connected, oriented $n$-manifolds. Since $H_n(M)=H_n(N)$ is infinite cyclic, the induced homomorphism $f_*:H_n(M)\to H_n(N)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$.

In the special case when we are computing the degree for a map $f:S^n\to S^n$, by excision we get
$$\deg(f) = \sum_{x_i\in f^{-1}(y)} \deg\left(H_n(U_i,U_i-x_i)\tov{f_*} H_n(V,V-y)\right),$$ for any $y\in Y$, some neighborhood $V$ of $y$, and preimages $U_i$ of $V$. This is called the local degree of $f$.

Geometry

Let $M$ be a smooth $n$-manifold. Recall $\Omega_M^r$ is the space of $r$-forms on $M$ and $d^r:\Omega^r_M\to \Omega^{r+1}_M$ is the differential map. Also recall the de Rham cohomology groups $H^r(M) = \text{ker}(d^r)/\text{im}(d^{r-1})$.

Definition: An $n$-manifold $M$ is orientable if it has a nowhere-zero $n$-form $\omega\in \Omega^n_M$. A choice of $\omega$ is called an orientation of $M$.

We also have a map $H^n(M)\to \R$, given by $\alpha\mapsto \int_M\alpha$, where the integral is normalized by the volume of $M$, so that integrating 1 across $M$ gives back 1. It is immediate that this doesn't make sense when $M$ is not compact, but when $M$ is compact and orientable, we get that $H^n(M)\neq0$. Indeed, if $\eta\in \Omega^{n-1}_M$ with $d\eta =\omega$, by Stokes' theorem we have
$$\int_M\omega = \int_Md\eta = \int_{\dy M}\eta = \int_\emptyset\eta = 0,$$
as $M$ has no boundary (since it is a manifold). But $\omega$ is nowhere-zero, meaning the first expression on the left cannot be zero. Hence $\omega$ is not exact and is a non-trivial element of $H^n(M)$.

Theorem: Let $M$ be a smooth, compact, orientable manifold of dimension $n$. Then $H^n(M)$ is one-dimensional.

Proof: The above discussion demonstrates that $\dim(H^n(M))\>1$. We can get an upper bound on the dimension by noting that the space of $n$-forms on $M$, given by $\Omega^n_M = \bigwedge^n(TM)^*$, has elements described by $dx_{i_1}\wedge \cdots \wedge dx_{i_n}$, with $\{i_1,\dots,i_n\}\subset \{1,\dots,n\}$. By rearranging the order of the $dx_{i_j}$, every element looks like $\alpha dx_1\wedge \cdots\wedge dx_n$ for some real number $\alpha$. Hence $\dim(\Omega^n_M) \leqslant 1$, so $\dim(H^n(M))$ is either 0 or 1. Therefore $\dim(H^n(M))=1$. $\square$

Definition: Let $f:M\to N$ be a map of smooth, compact, oriented manifolds of dimension $n$. Since $H^n(M)$ and $H^n(N)$ are 1-dimensional, the induced map $f^*:H^n(N)\to H^n(M)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$. Equivalently, for any $\omega\in \Omega^n_N$,
$$\int_M f^*\omega = d\int_N\omega$$

References: Hatcher (Algebraic Topology, Chapters 2, 3.3), Lee (Introduction to Smooth Manifolds, Chapter 17)

The Grassmannian is a complex manifold

 Lecture topic

Let $Gr(k,\C^n)$ be the space of $k$-dimensional complex subspaces of $\C^n$, also known as the complex Grassmannian. We will show that it is a complex manifold of dimension $k(n-k)$. Thanks to Jinhua Xu and professor Mihai Păun for explaining the details.

To begin, take $P\in Gr(k,\C^n)$ and an $n-k$ subspace $Q$ of $\C^n$, such that $P\cap Q = \{0\}$. Then $P\oplus Q = \C^n$, so we have natural projections

A neighborhood of $P$, depending on $Q$ may be described as $U_Q = \{S\in Gr(k,\C^n)\ :\ S\cap Q = \{0\}\}$. We claim that $U_Q \cong \Hom(P,Q)$. The isomorphism is described by
$$\begin{array}{r c c c l} \Hom(P,Q) & \to & U_Q & \to & \Hom(P,Q), \\ A & \mapsto & \{v+Av\ :\ v\in P\}, \\ & & S & \mapsto & \left(\pi_Q|_S\right) \circ \left(\pi_P|_S\right)^{-1}. \end{array}$$
The map on the right, call it $\varphi_Q$, is also the chart for the manifold structure. The idea of decomposing $\C^n$ into $P$ and $Q$ and constructing a homomorphism from $P$ to $Q$ may be visualized in the following diagram.

Then $\Hom(P,Q) \cong \Hom(\C^k,\C^{n-k})\cong \C^{k(n-k)}$, so $Gr(k,\C^n)$ is locally of complex dimension $k(n-k)$. To show that there is a complex manifold structure, we need to show that the transition functions are holomorphic. Let $P,P'\in Gr(k,\C^n)$ and $Q,Q'\in Gr(n-k,\C^n)$ such that $P\cap Q = P'\cap Q' = \{0\}$. Let $X\in \Hom(P,Q)$ such that $X\in \varphi_Q(U_Q\cap U_{Q'})$, with $\varphi_Q(S)=X$ and $\varphi_{Q'}(S)=X'$ for some $S\in U_Q\cap U_{Q'}$. Define $I_X(v) = v+Xv$, and note the transition map takes $X$ to
$$\begin{align*} X' & = \varphi_{Q'}\circ \varphi_Q^{-1}(X) & (\text{definition}) \\ & = \varphi_{Q'}(S) & (\text{assumption})\\ & = \left(\pi_{Q'}|_S\right)\circ \left(\pi_{P'}|_S\right)^{-1} & (\text{definition}) \\ & = \left(\pi_{Q'}|_S\right)\circ I_X\circ I_X^{-1}\circ \left(\pi_{P'}|_S\right)^{-1} & (\text{creative identity}) \\ & = \left(\pi_{Q'}|_S\circ I_X\right)\circ \left(\pi_{P'}|_S\circ I_X\right)^{-1} & (\text{redistribution}) \\ & = \left(\pi_{Q'}|_P +\pi_{Q'}|_Q\circ X\right) \circ \left(\pi_{P'}|_P +\pi_{P'}|_Q\circ X\right). & (\text{definition}) \end{align*}$$
At this last step we have compositions and sums of homomorphisms and linear maps, which are all holomorphic. Hence the transition functions of $Gr(k,\C^n)$ are holomorphic, so it is a complex manifold.

Complexes and their homology

 Preliminary exam prep

Here I'll present complexes from the most restrictive to the most general. Recall the standard $n$-simplex is
$$\Delta^n = \{x\in \R^{n+1}\ :\ \textstyle\sum x_i = 1, x_i\>0\}.$$

Definition: Let $V$ be a finite set. A simplicial complex $X$ on $V$ is a set of distinct subsets of $V$ such that if $\sigma\in X$, then all the subsets of $\sigma$ are in $X$.

Every $n$-simplex in a simplicial complex is uniquely determined by its vertices, hence no pair of lower dimensional faces of a simplex may be identified with each other.

Definition: Let $A,B$ be two indexing sets. A $\Delta$-complex (or delta complex) $X$ is
$$X = \left.\bigsqcup_{\alpha\in A} \Delta^{n_\alpha}_\alpha \right/\left\{\mathcal F_{\beta}^{k_\beta}\right\}_{\beta\in B}\ , \hspace{1cm} \mathcal F_\beta^{k_\beta} = \{\Delta_1^{k_\beta},\dots,\Delta_{m_\beta}^{k_\beta}\},$$
such that if $\sigma$ appears in the disjoint union, all of its lower dimensional faces also appear. The identification of the $k$-simplices in $\mathcal F^k$ is done in the natural (linear) way, and restricting to identified faces gives the identification of the $\mathcal F^{k-1}$ where the faces appear.

To define simplicial homology of a simplicial or $\Delta$-complex $X$, fix an ordering of the set of 0-simplices (which gives an ordering of every $\sigma\in X$), define $C_k$ to be the free abelian group generated by all $\sigma\in X$ of dimension $k$ (defined by $k+1$ 0-simplices), and define a boundary map
$$\begin{array}{r c l} \partial_k\ :\ C_k & \to & C_{k-1}, \\\ [v_0,\dots,v_k] & \mapsto & \sum_{i=0}^k(-1)^i[v_0,\dots,\widehat{v_i},\dots,v_k]. \end{array}$$
Then $H_k(X):= \text{ker}(\partial_k)/\text{im}(\partial_{k+1})$.

Recall the standard $n$-cell is $e^n = \{x\in \R^n\ :\ | x| \leqslant 1\}$, also known as the $n$-disk or $n$-ball.


Definition: Let $X_0$ be a finite set. A cell complex (or CW complex) is a collection $X_0,X_1,\dots$ where
$$X_k := \left.X_{k-1}\bigsqcup_{\alpha\in A_k} e^k_\alpha \right/\left\{\partial e^k_\alpha\sim f_{k,\alpha}(\dy e^k_\alpha)\right\}_{\alpha\in A_k},$$
where the $f_{k,\alpha}$ describe how to attach $k$-cells to the $(k-1)$-skeleton $X_{k-1}$, for $k\>1$. $X_k$ may also be described by pushing out $e^k\sqcup_{\dy e^k}X_{k-1}$. Note that $\dy e^k = S^{k-1}$, the $(k-1)$-sphere.

To define cellular homology, we need more tools (relative homology and excision) that require a blog post of their own.

References: Hatcher (Algebraic topology, Chapter 2.1)

Classical Lie groups

 Lecture topic

A Lie group $G$ is both a group and a manifold, with a smooth map $G\times G\to G$, given by $(g,h)\mapsto gh^{-1}$. The Lie algebra $\mathfrak g$ of $G$ is the tangent space $T_eG$ of $G$ at the identity.

We distinguish between real and complex Lie groups by saying that the base manifold is either real or complex analytic, respectively.

Example: Here are some examples of classical Lie groups and their dimension:
$$\begin{array}{r c r c l} \text{general linear group} & n^2 & GL(n) & = & \{n\times n\ \text{matrices with non-zero determinant}\} \\ \text{special linear group} & n^2-1 & SL(n) & = & \{M\in GL(n)\ :\ \det(M)=1\} \\ \text{orthogonal group} & n(n-1)/2 & O(n) & = & \{M\in GL(n)\ :\ MM^t = I\} \\ \text{special orthogonal group} & n(n-1)/2 & SO(n) & = & \{M\in O(n)\ :\ \det(M)=1\} \\ \text{unitary group} & n^2 & U(n) & = & \{M\in GL(n,\textbf{C})\ :\ MM^* = I\} \\ \text{special unitary group} & n^2-1 & SU(n) & = & \{M\in U(n)\ :\ \det(M)=1\} \\ \text{symplectic group} & n(2n+1) & Sp(n) & = & \{n\times n\ \text{matrices}:\ \omega(Mx,My)=\omega(x,y)\} \end{array}$$
For the symplectic group, the skew-symmetric bilinear form $\omega$ is defined as
$$\omega(x,y) = \sum_{i=1}^n x_iy_{i+n} - y_ix_{i+n} = \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} x\cdot y,$$
where $\cdot$ is the regular dot product (a symmetric bilinear form). Also note that the unitary group is a real Lie group - real because there is no holomorphic map $G\times G\to G$ as would be necessary, so we view the entries of a matrix in $U(n)$ in terms of its real and imaginary parts. Hence the dimension indicated above is real dimension.

References: Kirillov Jr (An introduction to Lie groups and Lie algebras, Chapter 2)

The real and complex Jacobian

 Lecture topic

Let $f:\C^n\to \C^n$ be a holomorphic function. We will show that the the real Jacobian is the square of the complex Jacobian. Write $f = (f_1,\dots,f_n)$ with $f_i = u_i+\img v_i$, where the $u_i$ are functions of the $z_j = x_j+\img y_j$. By the Cauchy-Riemann equations
$$\frac{\dy u_i}{\dy x_j} = \frac{\dy v_i}{\dy y_j} \hspace{1cm} \text{and} \hspace{1cm} \frac{\dy u_i}{\dy y_j} = -\frac{\dy v_i}{\dy x_j}$$
and expanding, we have that
$$\begin{align*} \frac{\dy f_i}{\dy z_j} & = \frac 12\left(\frac{\dy f_i}{\dy x_j} - \img \frac{\dy f_i}{\dy y_j}\right) \\ & = \frac12\left(\frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j} - \img\left(\frac{\dy u_i}{\dy y_j} + \img \frac{\dy v_i}{\dy y_j}\right)\right) \\ & = \frac12 \left(\frac{\dy u_i}{\dy x_j}+ \frac{\dy v_i}{\dy y_j} + \img \left(\frac{\dy v_i}{\dy x_j} - \frac{\dy u_i}{\dy y_j}\right)\right) \\ & = \frac{\dy u_i}{\dy x_j} + \img \frac{\dy v_i}{\dy x_j}. \end{align*}$$
The complex Jacobian of $f$ is $J_\C f$ (or its determinant), with entries
$$(J_\C f)_{i,j} = \frac{\dy f_i}{\dy z_j},$$
and the real Jacobian of $f$ is $J_\R f$ (or its determinant), with entries
$$\begin{align*} \begin{bmatrix} (J_\R f)_{2i-1,2j-1} & (J_\R f)_{2i-1,2j} \\ (J_\R f)_{2i,2j-1} & (J_\R f)_{2i,2j} \end{bmatrix} & = \begin{bmatrix} \displaystyle \frac{\dy u_i}{\dy x_j} & \displaystyle \frac{\dy u_i}{\dy y_j} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{R_{2i-1} + \img R_{2i} \to R_{2i-1}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & \displaystyle \img\frac{\dy f}{\dy z} \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \frac{\dy v_i}{\dy y_j} \end{bmatrix} \\ & \tov{C_{2j} - \img C_{2j-i} \to C_{2j}} \begin{bmatrix} \displaystyle \frac{\dy f_i}{\dy z_j} & 0 \\ \displaystyle \frac{\dy v_i}{\dy x_j} & \displaystyle \overline{\frac{\dy f_i}{\dy z_j}} \end{bmatrix}, \end{align*}$$
where the row and column operations have been performed for all rows $2i$ and all columns $2j$. Moving all the odd-indexed columns to the left and all odd-indexed rows to the top, we get that
$$J_\R f \simeq \begin{bmatrix} A & 0 \\ * & B \end{bmatrix} \hspace{1cm} \text{with} \hspace{1cm} A_{i,j} = \frac{\dy f_i}{\dy z_j},\ \ \ B_{i,j} = \overline{\frac{\dy f_i}{\dy z_j}}.$$
Since the number of operations to switch the columns is the same as the number of operations to switch the rows, the sign of the determinant of $J_\R f$ will not change. That is,
$$\det(J_\R f) = \det(A)\det(B) = \det(J_\C f) \overline{\det( J_\C f)} = |\det(J_\C f)|^2.$$