commit fe3576e04e90396a6455ff69235e8e5cdf40b125
parent c5db8ffc0f78e8b8bcfe856cb4a52e8948438890
Author: miksa234 <milutin@popovic.xyz>
Date: Sun, 16 May 2021 16:28:49 +0200
environment change from align* to align
Diffstat:
16 files changed, 133 insertions(+), 133 deletions(-)
diff --git a/pdfs/pres.pdf b/pdfs/pres.pdf
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diff --git a/pdfs/week1.pdf b/pdfs/week1.pdf
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diff --git a/pdfs/week2.pdf b/pdfs/week2.pdf
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diff --git a/pdfs/week4.pdf b/pdfs/week4.pdf
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diff --git a/src/week1.tex b/src/week1.tex
@@ -200,9 +200,9 @@ Under the pointwise product:
\subsubsection{Matrix Algebras}
\begin{definition}
A \textit{(complex) matrix algebra} A is a direct sum, for $n_i, N \in \mathbb{N}$.
- \begin{align*}
+ \begin{align}
A = \bigoplus _{i=1}^{N} M_{n_i}(\mathbb{C})
- \end{align*}
+ \end{align}
The involution is the hermitian conjugate, a $*$ algebra with involution is referred to as
a matrix algebra
\end{definition}
@@ -227,18 +227,18 @@ finite dimensional inner product space $H$ (finite Hilbert-spaces), with inner p
$(\cdot,\cdot)\rightarrow \mathbb{C}$. $L(H)$ is the $*$-algebra of operators on $H$
with product given by composition and involution given by the adjoint, $T \mapsto T^*$.
$L(H)$ is a \textit{normed vector space} with
-\begin{align*}
+\begin{align}
&\|T\|^2 = \text{sup}_{h \in H}\{(Th,Th): (h,h) \leq 1\} \hspace{0.1\textwidth} T \in L(H) \\
&\|T\| = \text{sup}\{\sqrt{\lambda}: \lambda \text{ eigenvalue of } T\}
-\end{align*}
+\end{align}
\begin{definition}
The \textit{representation} of a finite dimensional $*$-algebra A is a pair $(H, \pi)$.
$H$ is a finite, dimensional inner product space and $\pi$ is a $*$-\textit{algebra map}
- \begin{align*}
+ \begin{align}
\pi:A\ \rightarrow \ L(H)
- \end{align*}
+ \end{align}
\end{definition}
\begin{definition}
$(H, \pi)$ is called \textit{irreducible} if:
@@ -260,9 +260,9 @@ Examples for reducible and irreducible representations
\begin{definition}
Let $(H_1, \pi _1)$ and $(H_2, \pi _2)$ be representations of a $*$-algebra $A$. They are called
\textit{unitary equivalent} if there exists a map $U: H_1 \rightarrow H_2$ such that.
- \begin{align*}
+ \begin{align}
\pi _1(a) = U^* \pi _2(a) U
- \end{align*}
+ \end{align}
\end{definition}
\begin{question}
@@ -285,9 +285,9 @@ More on that in later chapters.
\textbf{
Given $(H, \pi)$ of a $*$-algebra $A$, the \textbf{commutant} $\pi (A)'$ of $\pi (A)$ is defined as a set
of operators in $L(H)$ that commute with all $\pi (a)$
- \begin{align*}
+ \begin{align}
\pi (A)' = \{T \in L(H):\pi (a)T = T\pi (a) \;\;\; \forall a\in A\}
- \end{align*}
+ \end{align}
\begin{enumerate}
\item Show that $\pi (A)'$ is a $*$-algebra.
\item Show that a representation $(H, \pi)$ of $A$ is irreducible iff the commutant $\pi (A)'$
@@ -354,26 +354,26 @@ of matrix algebras, to preserve general non-commutivity of matrices.
\begin{enumerate}
\item \textit{left} A-module is a vector space $E$, that carries a left representation of $A$,
that is $\exists$ a bilinear map $\gamma: A \times E \rightarrow E$ with
- \begin{align*}
+ \begin{align}
(a_1a_2)\cdot e = a_1 \cdot (a_2 \cdot e);\;\;\; a_1, a_2 \in A, e \in E
- \end{align*}
+ \end{align}
\item \textit{right} B-module is a vector space $F$, that carries a right representation of $A$,
that is $\exists$ a bilinear map $\gamma: F \times B \rightarrow F$ with
- \begin{align*}
+ \begin{align}
f \cdot (b_1b_2)= (f \cdot b_1) \cdot b_2;\;\;\; b_1, b_2 \in B, f \in F
- \end{align*}
+ \end{align}
\item \textit{left} A-module and \textit{right} B-module is a \textit{bimodule}, a vector space $E$
satisfying
- \begin{align*}
+ \begin{align}
a \cdot (e \cdot b)= (a \cdot e) \cdot b;\;\;\; a \in A, b \in B, e \in E
- \end{align*}
+ \end{align}
\end{enumerate}
\end{definition}
Notion of A-\textbf{module homomorphism} as linear map $\phi: E\rightarrow F$ which respects the
representation of A, e.g. for left module.
-\begin{align*}
+\begin{align}
\phi (ae) = a \phi (e); \;\;\; a \in A, e \in E.
-\end{align*}
+\end{align}
Remark on the notation
\begin{itemize}
\item ${}_A E$ left $A$-module $E$;
@@ -389,15 +389,15 @@ Remark on the notation
Not quite sure but \\
$a \in A$, $h_1, h_2 \in H$, we know $\pi (a) = T \in L(H)$ than
- \begin{align*}
+ \begin{align}
\langle \pi (a) h_1, \pi (a) h_2\rangle = \langle T h_1, T h_2\rangle = \langle T^*T h_1, h_2\rangle = \langle h_1, h_2\rangle
- \end{align*}
+ \end{align}
Or maybe this \\
If $_A H$ than $(a_1a_2) h = a_1 (a_2 h)$ for $a_1, a_2 \in A$ and $h \in H$.\\
Then we take the representation of an $a \in A$, $\pi (a)$:
- \begin{align*}
+ \begin{align}
(\pi(a_1)\pi(a_2))h = \pi(a_1)(\pi(a_2) h) = (T_1T_2) h = T_1 (T_2 h)
- \end{align*}
+ \end{align}
For $T_1, T_2 \in L(H)$, which operate naturally from the left on $h$.
\end{MyExercise}
diff --git a/src/week2.tex b/src/week2.tex
@@ -89,10 +89,10 @@
\begin{definition}
Let $A$ be an algebra, $E$ be a \textit{right} $A$-module and $F$ be a \textit{left} $A$-module.
The \textit{balanced tensor product} of $E$ and $F$ forms a $A$-bimodule.
- \begin{align*}
+ \begin{align}
E \otimes _A F := E \otimes F / \left\{\sum _i e_i a_i \otimes f_i - e_i \otimes a_i f_i : \;\;\;
a_i \in A,\ e_i \in E,\ f_i \in F \right\}
- \end{align*}
+ \end{align}
\end{definition}
Note $/$ denotes the quotient space. So $\otimes _A$ takes two left/right modules and makes a
bimodule with the help the tensor product of the two modules and the quotient space that takes
@@ -105,12 +105,12 @@ are duplicates.
\item an $B$-valued \textit{inner product} $\langle \cdot,\cdot\rangle_E: E\times E \rightarrow B$
\end{itemize}
$\langle \cdot,\cdot\rangle_E$ needs to satisfy the following for $e, e_1, e_2 \in E,\ a \in A$ and $b \in B$.
-\begin{align*}
+\begin{align}
\langle e_1, a\cdot e_2\rangle_E &= \langle a^*\cdot e_1, e_2\rangle_E \;\;\;\; & \text{sesquilinear in $A$}\\
\langle e_1, e_2 \cdot b\rangle_E &= \langle e_1, e_2\rangle_E b \;\;\;\; & \text{scalar in $B$} \\
\langle e_1, e_2\rangle_E &= \langle e_2,e_1\rangle^*_E \;\;\;\; & \text{hermitian} \\
\langle e, e\rangle_E &\ge 0 \;\;\;\; & \text{equality holds iff $e=0$}
-\end{align*}
+\end{align}
\end{definition}
@@ -140,9 +140,9 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\textbf{
Show that the $A-A$ bimodule given by $A$ is in $KK_f(A,A)$ by taking the following inner product
$\langle \cdot,\cdot\rangle_A:A \times A \rightarrow A$:
- \begin{align*}
+ \begin{align}
\langle a, a\rangle_A = a^*a' \;\;\;\; a,a'\in A
- \end{align*}
+ \end{align}
\label{exercise: inner-product}
}\newline
@@ -160,9 +160,9 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
From it we can construct a Hilbert bimodule $E_{\phi} \in KK_f(A, B)$ in the following way.
We let $E_{\phi}$ be $B$ in the vector space sense and an inner product from the above
Exercise \ref{exercise: inner-product}, with $A$ acting on the left with $\phi$.
- \begin{align*}
+ \begin{align}
a\cdot b = \phi(a)b \;\;\;\; a\in A, b\in E_{\phi}
- \end{align*}
+ \end{align}
\end{example}
@@ -171,13 +171,13 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\begin{definition}
Let $E \in KK_f(A, B)$ and $F \in KK_F(B, D)$ the \textit{Kasparov product} is defined as
with the balanced tensor product
- \begin{align*}
+ \begin{align}
F \circ E := E \otimes _B F
- \end{align*}
+ \end{align}
Such that $F\circ E \in KK_f(A,D)$ with a $D$-valued inner product.
- \begin{align*}
+ \begin{align}
\langle e_1 \otimes f_1, e_2 \otimes f_2\rangle _{E\otimes _B F} = \langle f_1,\langle e_1, e_2\rangle _E f_2\rangle _F
- \end{align*}
+ \end{align}
\end{definition}
\begin{question}
@@ -198,10 +198,10 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\item $E_{\text{id}_A} \simeq A \in KK_f(A,A)$
\item for $*$-algebra homomorphism $\phi: A \rightarrow B$ and $\psi: B \rightarrow C$ we have
an isomorphism
- \begin{align*}
+ \begin{align}
E_{\psi} \circ E_{\phi}\ \equiv\ E_{\phi} \otimes _B E_{\psi}\ \simeq\
E_{\psi \circ \phi} \in KK_f(A,C)
- \end{align*}
+ \end{align}
\end{enumerate}
}
\begin{enumerate}
@@ -242,9 +242,9 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\begin{definition}
Let $A$, $B$ be \textit{matrix algebras}. They are called \textit{Morita equivalent} if there
exists an $E \in KK_f(A, B)$ and an $F \in KK_f(B, A)$ such that:
- \begin{align*}
+ \begin{align}
E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B
- \end{align*}
+ \end{align}
Where $\simeq$ denotes the isomorphism between Hilbert bimodules, note that $A$ or $B$ is a bimodule by
itself.
\end{definition}
@@ -270,9 +270,9 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
standard $\mathbb{C}$ inner product.\\
On the other hand let $F = \mathbb{C}^n$, which is a $(\mathbb{C}, M_n(\mathbb{C}))$ Hilbert
bimodule by right matrix multiplication with $M_n(\mathbb{C})$ valued inner product:
- \begin{align*}
+ \begin{align}
\langle v_1, v_2\rangle =\bar{v_1}v_2^t \;\; \in M_n(\mathbb{C})
- \end{align*}
+ \end{align}
Now we take the Kasparov product of $E$ and $F$:
\begin{itemize}
\item $F\circ E\ =\ E\otimes _{\mathbb{C}}F\ \;\;\;\;\;\; \simeq \ M_n(\mathbb{C})$
@@ -287,13 +287,13 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\end{theorem}
\begin{proof}
Let $A$, $B$ be \textit{Morita equivalent}. So there exists $_A E_B$ and $_B F_A$ with
- \begin{align*}
+ \begin{align}
E \otimes _B F \simeq A \;\;\; \text{and} \;\;\; F \otimes _A E \simeq B
- \end{align*}
+ \end{align}
Consider $[(\pi _B, H)] \in \hat{B}$ than we construct a representation of $A$,
- \begin{align*}
+ \begin{align}
\pi _A \rightarrow L(E \otimes _B H)\;\;\; \text{with} \;\;\; \pi _A(a) (e \otimes v) = a e \otimes w
- \end{align*}
+ \end{align}
\begin{question}
Is $E \simeq H$ and $F \simeq W$? \\
Not in particular, there is a theorem that all infinite dimensional Hilbert spaces are isomorphic.
@@ -303,9 +303,9 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
the same.
\end{question}
\textit{vice versa}, consider $[(\pi _A, W)] \in \hat{A}$ we can construct $\pi _B$
- \begin{align*}
+ \begin{align}
\pi _B: B \rightarrow L(F \otimes _A W) \;\;\; \text{and}\;\;\; \pi _B(b) (f\otimes w) = bf\otimes w
- \end{align*}
+ \end{align}
These maps are each others inverses, thus $\hat{A} \simeq \hat{B}$
\end{proof}
@@ -337,10 +337,10 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\begin{proof}
We know $\mathbb{C}^n$ is a irreducible representation of $A= M_n(\mathbb{C})$. Let $H$ be irreducible
and of dimension $k$, then we define a map
- \begin{align*}
+ \begin{align}
\phi : A\oplus...\oplus A &\rightarrow H^* \\
(a_1,...,a_k) &\mapsto e^1\circ a_1^t+...+e^k\circ a_k^t
- \end{align*}
+ \end{align}
With $\{e^1,...,e^k\}$ being the basis of the dual space $H^*$ and $(\circ)$ being the pre-composition
of elements in $H^*$ and $A$ acting on $H$. This forms a morphism of $M_n(\mathbb{C})$ modules,
provided a matrix $a \in A$ acts on $H^*$ with $v\mapsto v\circ a^t$ ($v\in H^*$).
@@ -353,24 +353,24 @@ We denote $KK_f(A,B)$ the set of all \textit{Hilbert bimodules} of $(A,B)$.
\begin{example}
Consider two matrix algebras $A$, and $B$.
- \begin{align*}
+ \begin{align}
A = \bigoplus ^N_{i=1} M_{n_i}(\mathbb{C}) \;\;\; B = \bigoplus ^M_{j=1} M_{m_j}(\mathbb{C})
- \end{align*}
+ \end{align}
Let $\hat{A} \simeq \hat{B}$ that implies $N=M$ and define $E$ with $A$ acting by block-diagonal
matrices on the first tensor and B acting in the same way on the second tensor. Define $F$ vice versa.
- \begin{align*}
+ \begin{align}
E:= \bigoplus _{i=1}^N \mathbb{C}^{n_i} \otimes \mathbb{C}^{m_i} \;\;\;
F:= \bigoplus _{i=1}^N \mathbb{C}^{m_i} \otimes \mathbb{C}^{n_i}
- \end{align*}
+ \end{align}
Then we calculate the Kasparov product.
- \begin{align*}
+ \begin{align}
E \otimes _B F &\simeq \bigoplus _{i=1}^N (\mathbb{C}^{n_i}\otimes\mathbb{C}^{m_i})
\otimes _{M_{m_i}(\mathbb{C})} (\mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i}) \\
&\simeq \bigoplus _{i=1}^N \mathbb{C}^{n_i}\otimes
\left(\mathbb{C}^{m_i}\otimes _{M_{m_i}(\mathbb{C})}\mathbb{C}^{m_i}\right)
\oplus \mathbb{C}^{n_i} \\
&\simeq \bigoplus _{i=1}^N \mathbb{C}^{m_i}\otimes\mathbb{C}^{n_i} \simeq A
- \end{align*}
+ \end{align}
and from $F \otimes _A E \simeq B$.
\end{example}
diff --git a/src/week3.tex b/src/week3.tex
@@ -139,34 +139,34 @@
\end{align}
To further examine the exponent we rewrite the expression and Taylor expand $ln(1+K)$
to the second of $K = e^{i\alpha_a X_a} e^{i\beta_b X_b} -1$
- \begin{align*}
+ \begin{align}
i\delta _a X_a =& ln(1 + K) = K - \frac{K^2}{2} + \cdots \\
\text{and}\;\;\; K =&\ e^{i\alpha_a X_a} e^{i\beta_b X_b} -1 \\
=&\ (1 + i\alpha _a X_a - \frac{1}{2}(\alpha _a X_a)^2 + \cdots) \\
\cdot&\ (1 + i\beta _b X_b - \frac{1}{2}(\beta _b X_b)^2 + \cdots) -1 \\
=&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
-&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 + \cdots
- \end{align*}
+ \end{align}
So:
- \begin{align*}
+ \begin{align}
i\delta _a X_a =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
-&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 \\
- +&\ \frac{1}{2}(\ai\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b)^2 \\
+ +&\ \frac{1}{2}(\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b)^2 \\
=&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
-&\ \frac{1}{2}(\alpha _a X_a)^2 - \frac{1}{2}(\beta _b X_b)^2 \\
+&\ \frac{1}{2}(\alpha _a X_a)^2 + \frac{1}{2}(\beta _b X_b)^2 \\
+& \frac{1}{2}\alpha _a X_a \beta _b X_b + \frac{1}{2}\beta _b X_b \alpha _a X_a
- \end{align*}
+ \end{align}
Because $X$'s are linear operators $\alpha _a X_a \beta _b X_b \neq \beta _b X_b \alpha _a X_a$.
These generators form an \textit{algebra under commutation} and we get
- \begin{align*}
+ \begin{align}
i\delta _a X_a =&\ i\alpha _a X_a + i\beta _b X_b - \alpha_a X_a \beta _b X_b \\
-&\ \frac{1}{2}[\alpha _a X_a, \beta _b X_b] + \cdots
- \end{align*}
+ \end{align}
Thus rewriting the equation gives us
- \begin{align*}
+ \begin{align}
[\alpha _a X_a, \beta _b X_b] = -2i(\delta _c -\alpha _c -\beta _c) X_c \cdots \equiv i\gamma _c X_c
- \end{align*}
+ \end{align}
Because this is true for all $\alpha$ and $\beta$, and considering the group closure, there exists some
\textit{real} $f_{abc}$ called the \textit{structure constant} satisfying.
\begin{equation}
diff --git a/src/week4.tex b/src/week4.tex
@@ -62,21 +62,21 @@ Advantages of the characters are:
\item characters are different for inequivalent irreducible representation,
say $D_a$ and $D_b$ then there is a orthogonality relation up to $N$
(number of elements in the group):
- \begin{align*}
+ \begin{align}
\frac{1}{N} \sum _{g\in G} \chi _{D_a}(g)^*\chi_{D_b}(g) = \delta _{ab}
- \end{align*}
+ \end{align}
\end{itemize}
Furthermore characters are a \textit{complete} basis for functions that are constant on the conjugacy
class. Suppose $F(g_1)$ is such a function. We can expand this function in terms of matrix elements
of irreducible representations.
- \begin{align*}
+ \begin{align}
F(g_1) &= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{jk} \\
&= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g^{-1}g_1g))_{jk} \\
&= \sum_{a,j,k,g,l,m}\frac{1}{n_a} c_{jk}^{a} (D_a(g^{-1}))_{jl}(D_a(g_1))_{lm}(D_a(g))_{mk} \\
&= \sum_{a,j,l}\frac{1}{n_a} c_{jk}^{a} (D_a(g_1))_{lm} \delta _{jk} \delta _{lm} \\
&= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} (D_a(g_1))_{ll} \\
&= \sum_{a,j,l}\frac{1}{n_a} c_{jj}^{a} \chi _a (g_1)
- \end{align*}
+ \end{align}
where $n_a$ denotes the dimension of the representation $D_a$.
\newline
@@ -114,12 +114,12 @@ First we will see what we can find just by looking at the symmetries of the syst
The 6 degrees of freedom means that the system can be described with a 6 dimensional space.
This is a tensor product of a 2 dimensional space of $x$ and $y$ coordinates and a 3 dimensional
space of the masses (blocks).
-\begin{align*}
+\begin{align}
(x_1, y_1, x_2, y_2,x_3, y_3)
-\end{align*}
+\end{align}
The 3 dimensional space has $S_3$ symmetry (Group of all permutations of a three-element set),
it can be represented with $D_3$ the dihedral group.
-\begin{align*}
+\begin{align}
D_3(e) =
\begin{pmatrix}
1 & 0 & 0 \\
@@ -154,9 +154,9 @@ it can be represented with $D_3$ the dihedral group.
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{pmatrix} \;\;\;\;
-\end{align*}
+\end{align}
The 2 dimensional space also transforms under $S_3$, under a representation $D_2$
-\begin{align*}
+\begin{align}
D_2(e) =
\begin{pmatrix}
1 & 0 \\
@@ -185,20 +185,20 @@ The 2 dimensional space also transforms under $S_3$, under a representation $D_2
\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
-\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
\end{pmatrix} \;\;\;\;
-\end{align*}
+\end{align}
Then $D_6$ is the tensor product of $D_3$ and $D_2$:
\begin{equation}
(D_6(g))_{i\mu k\nu}=(D_3(g))_{ij}(D_2(g)_{\mu \nu}
\end{equation}
For $a_2$ we would than have block matrices instead of $1$ in $D_3$:
-\begin{align*}
+\begin{align}
D_6(a_2) =
\begin{pmatrix}
0 & D_2(a_2) & 0 \\
0 & 0 & D_2(a_2) \\
D_2(a_2) & 0 & 0 \\
\end{pmatrix}
-\end{align*}
+\end{align}
All other elements follow accordingly.
Now $S_3$ has two 1 dimensional irreducible representations which are trivial because they map to the
@@ -285,7 +285,7 @@ And for $D_2$ we get:
\end{align}
And the nontrivial modes provided by $P_2$ can be calculated by including translation in $x$ and $y$
direction $T_x$ and $T_y$:
-\begin{align*}
+\begin{align}
T_x = \frac{1}{3}
\begin{pmatrix}
1 & 0 & 1 & 0 & 1 & 0 \\
@@ -305,7 +305,7 @@ direction $T_x$ and $T_y$:
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 \\
\end{pmatrix}
-\end{align*}
+\end{align}
To see the modes we move mass $3$ in the $y$ direction which is the vector $\begin{pmatrix}0&0&0&0&0&1\end{pmatrix}$ and its mode can be get by appying it to $P_2 -T_x -T_y$:
\begin{equation}
\begin{pmatrix}
@@ -401,21 +401,21 @@ mode from Eq. \ref{eig: 3}.
The physical approach would be to construct the lagrangian $\mathfrak{L} = T - V$. Were $T$ is
simply the kinetic energy of the system in $\eta = (x_1, y_1, x_2, y_2, x_3, y_3)$ coordinates and
for simplicity we set all masses to $m$.
-\begin{align*}
+\begin{align}
T = \frac{m}{2} \dot{\eta} _i \dot{\eta} ^i \;\;\;\; \text{for\footnote{Einstein Summation Convention}}
\;\; i = 1,\dots,6
-\end{align*}
+\end{align}
For $V$ the potential energy we have three springs and \textit{small oscillations around equilibrium},
two of them are the offset of the one to the angle
of $\theta = \pm \frac{\pi}{3}$, which is $\begin{pmatrix} cos(\theta) \\ sin(\theta)\end{pmatrix} =
\begin{pmatrix} \frac{1}{2} \\ \pm \frac{\sqrt{3}}{2}\end{pmatrix}$. Then $V$ is:
-\begin{align*}
+\begin{align}
V &= \frac{k}{2} U^i_{\; j} \eta _i \eta ^j \\
&= \frac{k}{2} \bigg((x_1 - x_2)^2 +(\frac{1}{2}(x_2-x_3 + \frac{\sqrt{3}}{2} (y_2 - y_3))^2
+(\frac{1}{2}(x_1-x_3) + \frac{\sqrt{3}}{2}(y_1 - y_3))^2 \bigg)
-\end{align*}
+\end{align}
Where $U$ is:
-\begin{align*}
+\begin{align}
U = \frac{1}{4}
\begin{pmatrix}
5 & \sqrt{3} & -4 & 0 & -1 & -\sqrt{3} \\
@@ -425,7 +425,7 @@ Where $U$ is:
-1 & -\sqrt{3} & -1 & \sqrt{3} & 2 & 0 \\
-\sqrt{3} & -3 & \sqrt{3} & -3 & 0 & 6
\end{pmatrix}
-\end{align*}
+\end{align}
The Euler-Lagrange equations then give us :
\begin{equation}
diff --git a/src/week5.tex b/src/week5.tex
@@ -176,14 +176,14 @@ The metric is then:
Compute the metric on the space of three points given by $d_{ij} =
\sup_{a\in A}\{|a(i) - a(j)|: ||[D, \pi(a)]|| \leq 1\}$ for the set of data
$A = \mathbb{C}^3$ acting in the defining representation $H = \mathbb{C}^3$, and
- \begin{align*}
+ \begin{align}
D =
\begin{pmatrix}
0 & d^{-1} & 0 \\
d^{-1} & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
- \end{align*}
+ \end{align}
for some $d \in \mathbb{R}$
}\newline
@@ -223,11 +223,11 @@ The metric is then:
}\newline
For Equation \ref{metric 1} set $i=j$ in \ref{ext metric}.
-\begin{align*}
+\begin{align}
d_{ii} &= \sup_{a \in A}\{|\text{Tr}(a(i)) - \text{Tr}((a(i))|: ||[D, a]|| \leq
1\big\} \\
&= \sup_{a \in A}\{0: ||[D, a]|| \leq 1\big\} = 0
-\end{align*}
+\end{align}
For Equation \ref{metric 2} obviously we have the commuting property of
addition.
\newline
@@ -295,44 +295,44 @@ Then there is a map $d:A\rightarrow \Omega _D ^1 (A)$, $d = [D, \cdot]$.
\begin{MyExercise}
\textbf{
Verify that 'd` is a derivation of the C* algebra
- \begin{align*}
+ \begin{align}
d(ab) = d(a)b + ad(b) \\
d(a^*) = -d(a)^*
- \end{align*}
+ \end{align}
}\newline
For the record $d(\cdot) = [D, \cdot]$, then we have
\begin{enumerate}
\item
- \begin{align*}
+ \begin{align}
d(ab) &= [D, ab] = [D, a]b + a[D,b]\\
&= d(a)b + ad(b)
- \end{align*}
+ \end{align}
\item
- \begin{align*}
+ \begin{align}
d(a^*) &= [D, a^*] = Da^* - a^*D \\
&=-(D^*a - aD^*) = -[D^*, a] \\
&= -d(a)^*
- \end{align*}
+ \end{align}
\end{enumerate}
\end{MyExercise}
\begin{MyExercise}
\textbf{
Verify that $\Omega _D^1 (A)$ is an $A$-bimodule by rewriting
}
- \begin{align*}
+ \begin{align}
a(a_k[D, b_k])b = \sum_k a'_k[D, b'_k] \;\;\;\; a'_k, b'_k \in A
- \end{align*}
+ \end{align}
\newline
Begin
- \begin{align*}
+ \begin{align}
a(a_k[D, b_k])b &= aa_k(Db_k - b_k D) b = \\
&= aa_k(Db_k b - b_k D b) = aa_k(Db_k b - b_k Db - b_kbD +b_kbD)=\\
&= aa_k(Db_kb-b_kbD + b_k b D - b_k D b) = \\
&= aa_k [D, b_kb] + aa_k b [D, b]=\\
&= \sum _k a_k' [D, b_k']
- \end{align*}
+ \end{align}
\end{MyExercise}
@@ -414,34 +414,34 @@ Some remarks
\newline
For symmetry we need
- \begin{align*}
+ \begin{align}
(A_1, H_1, D_1) \sim (A_2, H_2, D_2) \Leftrightarrow
(A_2, H_2, D_2) \sim (A_1, H_1, D_1)
- \end{align*}
+ \end{align}
because $U$ is unitary:
- \begin{align*}
+ \begin{align}
&U\pi_1(a)U^* = \pi_2(a) \;\;\; | \cdot U^*\boxdot U \\
&U^*U\pi_1(a)U^*U = \pi_1(a) = U^*\pi_2(a)U \\
- \end{align*}
+ \end{align}
The same with the symmetric operator $D$.
\newline
For transitivity we need
- \begin{align*}
+ \begin{align}
(A_1, H_1, D_1) &\sim (A_2, H_2, D_2) \;\;\; \text{and} \;\;\;
(A_2, H_2, D_2) \sim (A_3, H_3, D_3) \\
&\Rightarrow (A_1, H_1, D_1) \sim (A_3, H_3, D_3)
- \end{align*}
+ \end{align}
There are two unitary maps $U_{12}:H_1 \rightarrow H_2$ and
$U_{23}: H_2 \rightarrow H_3$ then
- \begin{align*}
+ \begin{align}
U_{23}U_{12} \pi_1(a) U^*_{12}U^*_{23} &= U_{23}
\pi_2(a) U_23^* \\
&= \pi_3(a) \\
U_{23}U_{12} D_1U^*_{12}U^*_{23} &= U_{23}
D_2 U_23^* \\
&= D_3
- \end{align*}
+ \end{align}
\end{MyExercise}
Extending the this relation we look again at the notion of equivalence from
@@ -478,13 +478,13 @@ associated with the derivation $d=[D, \cdot]$ and satisfying the
\nabla(ae) = \nabla(e)a + e \otimes [D, a] \;\;\;\;\; e\in E,\; a\in A
\end{equation}
Then $D'$ is well defined on $E \otimes _A H$:
-\begin{align*}
+\begin{align}
D'(ea \otimes \xi - e \otimes a \xi) &= D'(ea \otimes \xi) - D'(e
\otimes \xi) \\
&= ea\otimes D\xi + \nabla(ae) \xi - e \otimes D(a\xi ) - \nabla (e)a
\xi \\
&= 0.
-\end{align*}
+\end{align}
With the information thus far we can prove the following theorem
\begin{theorem}
If $(A, H, D)$ a finite spectral triple, $E \in KK_f(B, A)$.
@@ -500,7 +500,7 @@ With the information thus far we can prove the following theorem
theorem). The only thing left is to show that $D'$ is a symmetric
operator, this we can just compute. Let $e_1, e_2 \in E$ and $\xi _1,
\xi _2 \in H$ then
- \begin{align*}
+ \begin{align}
\langle e_1 \otimes \xi _1, D'(e_2 \otimes \xi_2)\rangle _{E\otimes _A H} &=
\langle \xi _1, \langle e_1, \nabla e_2\rangle _E \xi _2\rangle + \langle \xi _1 , \langle e_1, e_2\rangle _E D\xi
_2\rangle _H \\
@@ -509,7 +509,7 @@ With the information thus far we can prove the following theorem
&+ \langle D\xi _1,\langle e_1, e_2\rangle _E \xi _2\rangle _H - \langle \xi _1, [D, \langle e_1, e_2\rangle _E] \xi
_2 \rangle _H \\
&= \langle D'(e_1 \otimes \xi _1), e_2 \otimes \xi _2\rangle _{E \otimes _A H}
- \end{align*}
+ \end{align}
\end{proof}
\begin{MyExercise}
@@ -522,14 +522,14 @@ With the information thus far we can prove the following theorem
Both $\nabla$ and $\nabla'$ need to satisfy the Leiblitz rule, so
let's see if $\nabla - \nabla'$ does.
- \begin{align*}
+ \begin{align}
\nabla(ea)-\nabla'(ea)&=\nabla(e) + e\otimes[D, a]\\
&-(\nabla'(e)a + e\otimes[D',a])\\
&=\bar{\nabla}a + e\otimes(Da-aD-D'a+aD')\\
&=\bar{\nabla}a + e\otimes((D-D')a-a(D-D'))\\
&=\bar{\nabla}a + e\otimes[D', a]\\
&=\bar{\nabla}(ea)
- \end{align*}
+ \end{align}
Therefore $\nabla-\nabla'$ is a linear map.
\end{MyExercise}
@@ -544,9 +544,9 @@ With the information thus far we can prove the following theorem
when the connection is $d(\cdot)$.
\item Use 1) and 2) to show that any connection $\nabla:
A\rightarrow A\otimes_A \Omega_D^1(A)$ is given by
- \begin{align*}
+ \begin{align}
\nabla = d + \omega
- \end{align*}
+ \end{align}
where $\omega \in \Omega_D^1(A)$
\item Upon identifying $A\otimes_A H \simeq H$, what is the
difference operator $D'$ with the connection on $A$ given by
@@ -634,10 +634,10 @@ With the information thus far we can prove the following theorem
The operator $D_e$ between $\textbf{n}_i$ and $\textbf{n}_j$ add up to
$D_{ij}$
-\begin{align*}
+\begin{align}
D_{ij} = \sum\limits_{\substack{e = (\nu _1, \nu _2) \\ n(\nu _1) =
\textbf{n}_i \\ n(\nu _2) = \textbf{n}_j}} D_e
-\end{align*}
+\end{align}
\begin{theorem}
There is a on to one correspondence between finite spectral triples
diff --git a/src/week6.tex b/src/week6.tex
@@ -179,9 +179,9 @@ vector space with the opposite product
is given by right matrix multiplication.
}\newline
- \begin{align*}
+ \begin{align}
a^\circ \xi = J a^* J^{-1}\xi = Ja^* \xi^* = J\xi a=\xi^* a
- \end{align*}
+ \end{align}
\end{MyExercise}
\begin{MyExercise}
\textbf{
@@ -230,13 +230,13 @@ vector space with the opposite product
for $S^{-1}$ if not then $\xi$ wouldn't exist. $S^{-1}(A*\xi) = S^{-1}(H) = H$.
\item Since $S$ is bijective then $\Delta ^{1/2}$ and $J$ need to be bijective.\\
Now let $\xi _1 , \xi _2 \in H$.\\
- \begin{align*}
+ \begin{align}
<J \xi _1 , J \xi _2 > &= < J^*J\xi_1 , \xi_2>^* =\\
&= <(\Delta ^{1/2})^* S^* S \Delta ^{1/2} \xi_1, \xi_2>^* = \\
&= <(SS^*)^{1/2}S^*S(SS^*) \xi_1, \xi_2>^* =\\
&= <(SS^*SS^*)^{1/2} \xi_1, \xi_2>^* = \\
&= <\xi _1, \xi_2>^* = <\xi_2 , \xi_1>.
- \end{align*}
+ \end{align}
\end{enumerate}
\end{MyExercise}
\subsection{Morphisms Between Finite Real Spectral Triples}
@@ -288,26 +288,26 @@ and linearity in $A$:
Straightforward show properties of the Hilbert bimodule and its $B^{\circ}$
valued inner product. Let $\bar{e}_1, \bar{e}_2 \in E^{\circ}$ and $a^\circ \in A,
b^\circ \in B$. \\
- \begin{align*}
+ \begin{align}
<\bar{e}_1, a^\circ \bar{e}_2> &= <\bar{e}_1, Ja^*J^{-1} \bar{e}_2>=\\
&= <\bar{e}_1 , J a^* e_2> = \\
&= <J^{-1} e_1, a^* e_2> =\\
& = <a^* e_1, e_2>= <J^{-1}(a^\circ)^* J e_1, e_2> = \\
& = <J^{-1} (a^\circ)^* \bar{e}_1, e_2> =\\
& = <(a^\circ)^* \bar{e}_1 , \bar{e}_2>.
- \end{align*}
+ \end{align}
Next $<\bar{e}_1, \bar{e}_2 b^\circ> = <\bar{e}_1, \bar{e_2}> b^\circ$.
- \begin{align*}
+ \begin{align}
<\bar{e}_1, \bar{e}_2 b^\circ> &= <\bar{e}_1, \bar{e}_2 Jb^*J^{-1}> =\\
&= <\bar{e}_1, \bar{e_2}> Jb^*J^{-1} = \\
&= <\bar{e}_1, \bar{e}_2> b^\circ.
- \end{align*}
+ \end{align}
Then:
- \begin{align*}
+ \begin{align}
(<\bar{e}_1, \bar{e}_2)>_{E^\circ})^* &= (<e_2, e_1>_E)^* =\\
&= <e_1, e_2>_E^* = <\bar{e}_2, \bar{e}_2>_{E^\circ}
- \end{align*}
+ \end{align}
And of course $<\bar{e}, \bar{e}> = <e, e> \geq 0$
\end{MyExercise}
@@ -356,14 +356,14 @@ with $b^\circ = J' b^* (J')^{-1}$, $b^* \in B$ action on $H'$.
Hagime:
- \begin{align*}
+ \begin{align}
&\text{For one:}\\
&\tau \circ \nabla(ae) = \bar{\nabla}(a\bar{e}) = \bar{\nabla}(a^* \bar{e})\\
&\text{For two:}\\
&\tau \circ \nabla(ae) = \tau(\nabla(e)a) + \tau \circ(e \otimes d(a))=\\
&=a^*\bar{\nabla}(\bar{e}) - d(a)^* \otimes \bar{e}. \\
&= a^*\bar{\nabla}(\bar{e}) + d(a^*) \otimes \bar{e}.
- \end{align*}
+ \end{align}
\end{MyExercise}
Then the connections
\begin{align}
@@ -400,18 +400,18 @@ Finally for the grading
\begin{proof}
The only thing left is to check if the $KO$-dimension is preserved,
for this we check if the $\epsilon$'s are the same.
- \begin{align*}
+ \begin{align}
&(J')^2 = 1 \otimes J^2 \otimes 1 = \epsilon\\
&J' \gamma '= \epsilon ''\gamma'J'
- \end{align*}
+ \end{align}
and for $\epsilon '$
- \begin{align*}
+ \begin{align}
J'D'(e_1 \otimes \xi \otimes \bar{e}_2)&=J'((\nabla e_1) \xi \otimes
\bar{e_2} + e_1 \otimes D\xi \otimes \bar{e}_2 + e_1 \otimes \xi (\tau
\nabla e_2))\\
&= \epsilon' D'(e_2 \otimes J\xi \otimes \bar{e}_2)\\
&= \epsilon' D'J'(e_1 \otimes \xi \bar{e}_2)
- \end{align*}
+ \end{align}
\end{proof}
\end{document}
diff --git a/src/week7.tex b/src/week7.tex
@@ -347,12 +347,12 @@ or a $-$ sign.
$v'$ implies that either $n(v) = n(v')$ or $m(v) = m('v)$ or both
\item of $e = (v_1, v_2) \in \Gamma^{(1)}$ edges with non-zero
operators $D_e$ and their adjoints $D_e^*$:
- \begin{align*}
+ \begin{align}
&D_e:\mathbb{C}^{n(v_1)} \rightarrow
\mathbb{C}^{n(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; m(v_1) = m(v_2)\\
&D_e:\mathbb{C}^{m(v_1)} \rightarrow
\mathbb{C}^{m(v_2)}\;\;\;\;\; &\text{if} \;\;\;\; n(v_1) = n(v_2)
- \end{align*}
+ \end{align}
\end{itemize}
Together with an involutive graph automorphism $j:\Gamma \Rightarrow
\Gamma$ such that the following conditions hold:
