From dfc89afc5455cbe3d5b88c7b6c22e8dc49960c0b Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 30 Sep 2025 22:22:19 -0500
Subject: [PATCH] updates

---
 .../Math401/Extending_thesis/Math401_R2.md    | 87 ++++++++++++++++++
 .../Math401/Extending_thesis/Math401_S4.md    | 90 ++++++++++++++++++-
 2 files changed, 175 insertions(+), 2 deletions(-)

diff --git a/content/Math401/Extending_thesis/Math401_R2.md b/content/Math401/Extending_thesis/Math401_R2.md
index 12b0cd7..9600b51 100644
--- a/content/Math401/Extending_thesis/Math401_R2.md
+++ b/content/Math401/Extending_thesis/Math401_R2.md
@@ -342,6 +342,93 @@ An atlas is said to be **smooth** if the transition maps $\phi_\alpha\circ \phi_
 
 A smooth manifold is a pair $(M,\mathcal{A})$ where $M$ is a topological manifold and $\mathcal{A}$ is a smooth atlas.
 
+#### Lie group
+
+Lie group is a group (satisfying group axioms: closure, associativity, identity, inverses) that is also a smooth manifold. with the operator $m:G\times G\to G$, and the inverse operation $i:G\to G$ that are both smooth.
+
+In short, a Lie group is a group that is also a smooth manifold with map $G\times G\to G$ given by $(g,h)\mapsto gh^-1$ that is smooth.
+
+<details>
+<summary>Example of Lie group</summary>
+
+The general linear group $GL(n,\mathbb{R})$ is the group of all $n\times n$ invertible matrices over $\mathbb{R}$.
+
+This is a Lie group since
+
+1. Multiplication is a smooth map $GL(n,\mathbb{R})\times GL(n,\mathbb{R})\to GL(n,\mathbb{R})$ since it is a polynomial map.
+2. Inverse is a smooth map $GL(n,\mathbb{R})\to GL(n,\mathbb{R})$ by cramer's rule.
+
+---
+
+If $G$ is a Lie group, then any open subgroup (with subgroup topology and open set in $G$) $H$ of $G$ is also a Lie group.  
+
+</details>
+
+#### Translation map on Lie group
+
+If $G$ is a Lie group, then the translation map $L_g:G\to G$ given by $L_g(h)=gh$ and $R_g:G\to G$ given by $R_g(h)=hg$ are both smooth and are diffeomorphisms on $G$.
+
+#### Derivation and tangent vectors
+
+The directional derivative of a geometric tangent vector $v_a\in \mathbb{R}^n_a$ yields a map $D_v\vert_a:C^\infty(\mathbb{R}^n)\to \mathbb{R}$ given by the formula
+
+$$
+D_v\vert_a(f)=D_v f(a)=\frac{d}{dt}\bigg\vert_{t=0}f(a+tv_a)
+$$
+
+Note that this is a linear over $\mathbb{R}$, and satisfies the product rule.
+
+$$
+D_v\vert_a(f\cdot g)=f(a)D_v\vert_a(g)+g(a)D_v\vert_a(f)
+$$
+
+We can generalize this representation to the following definition:
+
+If $a$ is a point of $\mathbb{R}^n$, then a **derivation at $a$** is a linear map $w:C^\infty(\mathbb{R}^n)\to \mathbb{R}$ such that it is linear over $\mathbb{R}$ and satisfies the product rule.
+
+$$
+w(f\cdot g)=w(f)\cdot g(a)+f(a)\cdot w(g)
+$$
+
+Let $T_a\mathbb{R}^n$ denote the set of all derivations of $C^\infty(\mathbb{R}^n)$ at $a$. So $T_a\mathbb{R}^n$ is a vector space over $\mathbb{R}$.
+
+$$
+(w_1+w_2)(f)=w_1(f)+w_2(f),\quad (cw)(f)=c(w(f))
+$$
+
+Some key properties are given below and check the proof in the book for details.
+
+1. If $f$ is a constant function, then $w(f)=0$.
+2. If $f(a)=g(a)=0$, then $w(f\cdot g)=0$.
+3. For each geometric tangent vector $v_a\in \mathbb{R}^n_a$, the map $D_v\vert_a:C^\infty(\mathbb{R}^n)\to \mathbb{R}$ is a derivation at $a$.
+4. The map $v_a\mapsto D_v\vert_a$ is an isomorphism of vector spaces from $\mathbb{R}^n_a$ to $T_a\mathbb{R}^n$.
+
+#### Tangent vector on Manifolds
+
+Let $M$ be a smooth manifold. Let $p\in M$. A **tangent vector to $M$ at $p$** is a derivation at $p$ if it satisfies:
+
+$$
+v(f\cdot g)=f(p)vg+g(p)vf\prod \text{ for all } f,g\in C^\infty(M)
+$$
+
+The set of all derivations of $C^\infty(M)$ at $p$ is denoted by $T_pM$ is called tangent space to $M$ at $p$. An element of $T_pM$ is called a tangent vector to $M$ at $p$.
+
+#### Tangent bundle
+
+We define the tangent bundle of $M$ as the disjoint union of all the tangent spaces:
+
+$$
+TM=\bigsqcup_{p\in M} T_pM
+$$
+
+We write the element in $TM$ as pair $(p,v)$ where $p\in M$ and $v\in T_pM$.
+
+The tangent bundle comes with a natural projection map $\pi:TM\to M$ given by $\pi(p,v)=p$.
+
+#### Vector field
+
+> CONTINUE HERE to study the importance of Lie algebra and Lie group for vector fields.
+
 ### Riemannian manifolds
 
 A Riemannian manifold is a smooth manifold equipped with a **Riemannian metric**, which is a smooth assignment of an inner product to each tangent space $T_pM$ of the manifold.
diff --git a/content/Math401/Extending_thesis/Math401_S4.md b/content/Math401/Extending_thesis/Math401_S4.md
index 9579beb..0394d79 100644
--- a/content/Math401/Extending_thesis/Math401_S4.md
+++ b/content/Math401/Extending_thesis/Math401_S4.md
@@ -134,6 +134,20 @@ $$
 |F(z)|^2\leq d_z \|F\|^2_{L^2(U,\alpha)}
 $$
 
+Suppose we have a sequence $F_n\in \mathcal{H}L^2(U,\alpha)$ such that $F_n\to F$, $F\in L^2(U,\alpha)$.
+
+Then $F_n$ is a cauchy sequence in $L^2(U,\alpha)$. So,
+
+$$
+\sup_{v\in V}|F_n(v)-F_m(v)|\leq \sqrt{d_z}\|F_n-F_m\|_{L^2(U,\alpha)}\to 0\text{ as }n,m\to \infty
+$$
+
+So the sequence $F_m$ converges locally uniformly to some limit function which must be $F$ ($\mathbb{C}^d$ is Hausdorff, unique limit point).
+
+Locally uniform limit of holomorphic functions is holomorphic. (Use Morera's Theorem to show that the limit is still holomorphic in each variable.) So the limit function $F$ is actually in $\mathcal{H}L^2(U,\alpha)$, which shows that $\mathcal{H}L^2(U,\alpha)$ is closed.
+
+which shows that $\mathcal{H}L^2(U,\alpha)$ is closed.
+
 </details>
 
 > [!TIP]
@@ -169,6 +183,45 @@ Let $\mathcal{H}L^2(U,\alpha)$ be a holomorphic space. The reproducing kernel of
    |F(z)|^2\leq K(z,z) \|F\|^2_{L^2(U,\alpha)}
    $$
 
+<details>
+<summary>Proof</summary>
+
+For part 1, By [Riesz Theorem](../../Math429/Math429_L27#theorem-642-riesz-representation-theorem), the linear functional evaluation at $z\in U$ on $\mathcal{H}L^2(U,\alpha)$ can be represented uniquely as inner product with some $\phi_z\in \mathcal{H}L^2(U,\alpha)$.
+
+$$
+F(z)=\langle F,\phi_z\rangle_{L^2(U,\alpha)}=\int_U F(w)\overline{\phi_z(w)} \alpha(w) dw
+$$
+
+And assume part 2 is true, then we have
+
+$K(z,w)=\overline{\phi_z(w)}$
+
+So part 1 is true.
+
+For part 2, we can use the same argument
+
+$$
+\phi_z(w)=\langle \phi_z,\phi_w\rangle_{L^2(U,\alpha)}=\overline{\langle \phi_w,\phi_z\rangle_{L^2(U,\alpha)}}=\overline{\phi_w(z)}
+$$
+
+... continue if needed.
+
+</details>
+
+#### Construction of reproducing kernel
+
+Let $\{e_j\}$ be any orthonormal basis of $\mathcal{H}L^2(U,\alpha)$. Then for all $z,w\in U$,
+
+$$
+\sum_{j=1}^{\infty} |e_j(z)\overline{e_j(w)}|<\infty
+$$
+
+and
+
+$$
+K(z,w)=\sum_{j=1}^{\infty} e_j(z)\overline{e_j(w)}
+$$
+
 ### Bargmann space
 
 The Bargmann spaces are the holomorphic spaces 
@@ -180,5 +233,38 @@ $$
 where
 
 $$
-\mu_t(z)=\text{ CONTINUE HERE }
-$$
\ No newline at end of file
+\mu_t(z)=(\pi t)^{-d}\exp(-|z|^2/t)
+$$
+
+> For this research, we can tentatively set $t=1$ and $d=2$ for simplicity so that you can continue to read the next section.
+
+#### Reproducing kernel for Bargmann space
+
+For all $d\geq 1$, the reproducing kernel of the space $\mathcal{H}L^2(\mathbb{C}^d,\mu_t)$ is given by
+
+$$
+K(z,w)=\exp(z\cdot \overline{w}/t)
+$$
+
+where $z\cdot \overline{w}=\sum_{k=1}^d z_k\overline{w_k}$.
+
+This gives the pointwise bounds
+
+$$
+|F(z)|^2\leq \exp(\|z\|^2/t) \|F\|^2_{L^2(\mathbb{C}^d,\mu_t)}
+$$
+
+For all $F\in \mathcal{H}L^2(\mathbb{C}^d,\mu_t)$, and $z\in \mathbb{C}^d$.
+
+> Proofs are intentionally skipped, you can refer to the lecture notes for details.
+
+#### Lie bracket of vector fields
+
+Let $X,Y$ be two vector fields on a smooth manifold $M$. The Lie bracket of $X$ and $Y$ is an operator $[X,Y]:C^\infty(M)\to C^\infty(M)$ defined by
+
+$$
+[X,Y](f)=X(Y(f))-Y(X(f))
+$$
+
+This operator is a vector field.
+