TITLE: Matrix inverse of block matrix $\begin{pmatrix} T & l \\l^*& S \end{pmatrix}$
QUESTION [1 upvotes]: Consider the special block matrix
$$A = \begin{pmatrix} T & l \\l^*& S \end{pmatrix},$$ where we assume that each entry is invertible and $l$ and $l^*$ denote a scalar (and its complex conjugate) times the identity, whereas $T$ and $S$ are full matrices of equal size.
Is there a simple formula for the matrix inverse $A^{-1}?$ I am aware there are general formulas for block matrix inverses, but I thought in this case, things should simplify?
Please let me know if you have any questions.

REPLY [2 votes]: Applying the block-inversion formula linked in the comments, we have
$$
\pmatrix{\mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}} = \\
\pmatrix{T^{-1}+|l|^2T^{-1}(S-|l|^2T^{-1})^{-1}T^{-1} &
-l\,T^{-1}(S-|l|^2 T^{-1})^{-1} \\
-l^*(S-|l|^2 T^{-1})^{-1}T^{-1} & (S-|l|^2T^{-1})^{-1}} = \\
\pmatrix{T^{-1}[1+|l|^2(TS-|l|^2)^{-1}] &
-l\,(ST-|l|^2 )^{-1} \\
-l^*(TS-|l|^2 )^{-1} & T(ST-|l|^2)^{-1}}
$$


TITLE: Explanation for Oxtoby's proof: a nonempty topological space $X$ is Baire iff player (I) has no winning strategy in the Choquet game
QUESTION [0 upvotes]: A nonempty topological space $X$ is a Baire space iff player I has no winning strategy in the Choquet game $G_X$.

Oxtoby's proof
I have several questions about this proof.
$(\Leftarrow)$

How can he assume the existence of a sequence $(G_n)$ of dense open sets?

Moreover, how can he assume that $\bigcap_n G_n \cap U_0 = \emptyset,$ if they're dense?

He takes a $V_0 \subseteq U_0$ and claims that $V_0 \cap G_0 \neq \emptyset$. But I thought we only knew that $\bigcap_n G_n \cap U_0 = \emptyset$? How can we know for sure that $G_0$ kills it?

$(\Rightarrow)$

What is it mean to show that $U_0$ is not "Baire"? I thought we had to show that $X$ is not Baire. I am confused about the logical implication too. If assuming player I has a winning strategy implies that $X$ is not Baire, then player I not having a winning strategy doesn't necessarily have to imply that $X$ is Baire?

Where he says that $\cup \mathcal {U}_p$ is dense in $U_n$, does that mean that the union of all possible responses $U_{n+1}$ is dense in $U_n$?

If $W_n = \cup \{U_n: (U_0, V_0, \ldots, U_n) \in S \}$ then isn't that the same as $\cup \mathcal{U}_p$ for $p = (U_0, V_0, \ldots, U_n)$? Then wouldn't we only have that $W_n$ is dense in $U_{n-1}$? How do we get that it is dense in $U_0$?

We claim that $\bigcap_n W_n = \emptyset.$ Otherwise, if $x \in \bigcap_n W_n$, there is unique $(U_0, V_0, U_1, V_1, \ldots ) \in [S]$ with $x \in U_n$ for each $n$, so $\bigcap_n U_n \neq \emptyset,$ a contradiction.

My question is how do we know that $(U_0, V_0, U_1, V_1, \ldots )$ is unique? I see how the $U_i$ are unique since they are disjoint, but what about the $V_i$?

Let, by an application of Zorn's Lemma, $\mathcal{V}_p$ be a maximal collection of nonempty open subsets $V_n \subseteq U_n$ such that $\{V_n^*: V_n \in \mathcal{V}_p \}$ is pairwise disjoint. Put in $S$ all $ (U_0, V_0, \ldots U_,n, V_n, U_{n+1})$ with $V_n \in \mathcal {V}_p$. Then $\mathcal{U}_p = \{U_{n+1}:(V_0, \ldots, U_n, V_n, U_{n+1}) \in S \} = \{V_n^*: V_n \in \mathcal{V}_p \}$ is a family of pairwise disjoint sets and $\bigcup \mathcal {U}_p$ is dense in $U_n$ by the maximality of $\mathcal {V}_p,$ since if $\bar{V_n} \subseteq U_n$ is nonempty, open and disjoint from $\bigcup \mathcal{U}_p$, then $\mathcal {V}_p \cup \{\bar{V}_n \}$ violates the maximality of $\mathcal {V}_p$.

I honestly have no idea what's going on here. Why distinguish between $\mathcal {V}_p$ and $\mathcal {U}_p$? What is $\bigcup \mathcal {U}_p$? Why do we need the sets to be pairwise disjoint? Why does the maximality of $\mathcal {V}_p$ imply that $\bigcup \mathcal {U}_p$ is dense and why do we need that this is dense?

REPLY [3 votes]: $\Leftarrow$:

Rearrange the statement as follows: since $X$ is not a Baire space, there exists a sequence $(G_n)$ of dense open sets in $X$ such that $\bigcap_n G_n$ is not dense in $X$ (we are simply denying one of the equivalent definitions of Baire Spaces). Hence we have a non-empty open set $U_0$ such that $\bigcap_n G_n \cap U_0 = \emptyset$.
see above
$G_0$ is a dense open set, as for all other $G_n$.

$\Rightarrow$:

Every open subset of a Baire space, is a Baire space. So if we prove that a non-empty open set $U_0 \subseteq X$ is not Baire, then we are actually proving that $X$ is not a Baire space.
What Oxtoby is trying to do is to provide a specific play of the game in which player I plays by $\sigma$ and such that player I winning ($\sigma$ is a winning strategy by assumption, so player I wins no matter what) implies that $U_0$ is not Baire. So the tree $S\subseteq \sigma$ he's building is a proper subset of all possible plays. So, no, $\bigcup \mathcal{U}_p$ being dense in $U_n$ means that the union of all responses $U_{n+1}$ *contained in $S$ * is dense in $U_n$, and those responses are a proper subset of all possible responses $U_{n+1}$ of player I.
No. $W_n$ is the union of all player 1 responses $U_n$ in the n-th stage of the plays. You can see it as $$W_n = \bigcup[\bigcup_{p = (U_0,V_0,\dots,U_{n-1}) \in S} \mathcal{U}_p]$$ It is easy to show that therefore $W_n$ is dense in $W_{n-1}$, and by a simple induction, proving that therefore $W_n$ is dense in $U_0 = W_0$ for all n.
The sequence is unique since not only $\mathcal{U}_p$ is a disjoint collection of open sets, but, again we can show this by a simple induction, that $$\bigcup_{p = (U_0,V_0,\dots,U_{n-1}) \in S} \mathcal{U}_p$$
is still a disjoint collection of open sets in $U_0$. Intuitively player 1's response in the subsequent stages of the games contained in $S$, are smaller and smaller disjoint open "refinements" of $U_0$. Keeping this in mind, it is easy to see that for all $n$, $\exists! s_n = (U_0,V_0,\dots,U_n) \in S$ such that $x \in U_n$ and that $s_{n} \subseteq s_{n+1}$ for all $n$ (hence we are defining a unique sequence by Dependent Choice).
Here we are simply constructing the subtree $S\subseteq \sigma$ we are working in from the start. What do we need? A subtree $S \subseteq \sigma$ such that $\mathcal{U}_p$ with $p = (U_0,\dots,U_n)$ is a disjoint collection of open sets such that $\bigcup \mathcal{U}_p$ is dense in $U_n$ for all $p \in S$. This kind of tree exists by Zorn's Lemma. See this, in this page the author explicitate this passage by a lemma. I find it quite clearer.


TITLE: Understanding the definition of stacks
QUESTION [16 upvotes]: First of all I should apologies if this question does not count as a research level one. I asked the same question on MathUnderflow and didn't receive any answer. Let me cross post (copy and paste) it here.
I am trying to understand the notion of sheaves and stacks. Intuitively, sheaves are bit easy to understand as a gluing of compatible families of sets assign to opens sets of a topological space. In other words, it is a contravariant functor $\mathcal{F}:\mathbf{Open}(X)^{\operatorname{op}}\to\mathbf{Set}$ such that
$$\mathcal{F}(U)=\lim\left(\prod_{i\in I}\mathcal{F}(U_i) \rightrightarrows \prod_{j,k\in I^2}\mathcal{F}(U_j\cap U_k)\right)$$
where $X$ is a topological space, $U$ is an open set of $X$ and $\{U_i\}_{i\in I}$ is any open cover of $U.$ Further, it is easy to imagine a sheaf as a étalé space over $X.$
Then I started reading about stacks using this notes and nlab as my primary sources. I learned that a stack is a contravariant functor $\mathcal{F}:\mathcal{C}^{\operatorname{op}}\to\mathbf{Grpd}$ satisfying a descent property and, categories fibered in groupoids over $\mathcal{C}$ is an intuitive way to think about stacks, where $\mathcal{C}$ is a site (category equipped with a coverage). Now I have following questions:

How can I understand the descent property for stack? To be more specific, how can triple fiber products (intersections) appear in the equalizer fork diagram?
What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks? Is this the correct analogue of étalé space of a stack?

REPLY [9 votes]: What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks?

A category fibered in groupoids over $\mathcal{C}$ is given by a functor $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ satisfying certain conditions (I am not writing the definition as I assume you already know what is a category fibered in groupoids); look at Definition 4.2 in the paper Orbifolds as stacks?
Put a Grothendieck topology on the category $\mathcal{C}$; considering it as a site.
Given an object $U$ of the category $\mathcal{C}$, we consider its fiber; a category, denoted by $\mathcal{F}(U)$, defined as
$$\text{Obj}(\mathcal{F}(U))=\{V\in \text{Obj}(\mathcal{F}):\pi_{\mathcal{F}}(V)=U\},$$
$$\text{Mor}_{\mathcal{F}(U)}(V_1,V_2)=\{(f:V_1\rightarrow V_2)\in \text{Mor}_{\mathcal{F}}(V_1,V_2):\pi_{\mathcal{F}}(f)=1_U\}.$$
Given a cover $\{U_\alpha\rightarrow U\}$ of the object $U$ (remember that we fixed a Grothendieck topology), we consider its descent category, denoted by $\mathcal{F}(\{U_\alpha\rightarrow U\})$.
An object of the category $\mathcal{F}(\{U_\alpha\rightarrow U\})$ is given by the following data:

for each index $i\in \Lambda$, an object $a_i$ in the category $\mathcal{F}(U_i)$,
for each pair of indices $i,j\in \Lambda$, an isomorphism $\phi_{ij}:pr_2^*(a_j)\rightarrow pr_1^*(a_i)$ in the category $\mathcal{F}(U_i\times_{U}U_j)$

satisfying appropriate cocycle condition.
Now, given a categroy fibered in groupoids $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$, an object $U$ of $\mathcal{C}$ and a cover $\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$ of $U$ in $\mathcal{C}$, there is an obvious functor $$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$
(which you might have guessed already but let me say that), at the level of objects $$a\mapsto ((a|_{U_\alpha}),(\phi_{ij}))$$

A category fibered in groupoids
  $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ is said to be a
  stack if, for each object $U$ of $\mathcal{C}$ and for each cover
  $\mathcal{U}(U)=\{U_\alpha\rightarrow U\}$, the functor
  $$p_{\mathcal{U}(U)}:\mathcal{F}(U)\rightarrow \mathcal{F}(\{U_\alpha\rightarrow U\})$$ is an equivalence of categories.

Now, you can ask what does equivalence of categories has anything to do with "sheaf like" properties? For a functor $\mathcal{D}\rightarrow \mathcal{C}$ to be an equivalence of categories, along other things, for each object $d\in \mathcal{D}$ we need an object $c\in \mathcal{C}$ such that, there is an isomorphism $F(c)\rightarrow d$.
Let $((a_\alpha),\{\phi_{\alpha\beta}\})$ be an object of $\mathcal{F}(\{U_\alpha\rightarrow U\})$. For this, by equivalence of categories, gives an element $a\in \mathcal{F}(U)$ that maps to $((a_\alpha),\{\phi_{\alpha\beta}\})$. That is, given an object $U$ of $\mathcal{C}$, an open cover $\{U_\alpha\rightarrow U\}$, for each collection of objects $\{a_\alpha\in \mathcal{F}(U_\alpha)\}$ that are compatible in some sense, there exists an object $a\in \mathcal{F}(U)$, such that, under appropriate restriction of $a$, you get the objects $a_{\alpha}$. This should remind the notion of sheaf on a topological space. This is how a stack is seen as a generalization of sheaf.
References :

Notes on Grothendieck topologies, fibered categories and descent theory.
Orbifolds as stacks?
How is a Stack the generalisation of a sheaf from a 2-category point of view?


text:
\section{An example: The H\'enon Map}  \label{sec:example}

The procedure for putting the mathematics of the previous sections
into practice on time-series data from a dynamical system is as
follows:

\begin{enumerate}
	\item Given a time series $\Gamma = \{x_0,\ldots,x_{T-1}\}
          \subset X$, which we assume lies near an
          invariant set $\Lambda\subseteq X$, select a set of
          landmarks, $L = \{l_0,\ldots,l_{\ell-1}\}$, that are evenly
          distributed across $\Lambda$ (cf.,
          page~\pageref{page:landmarks} and \App{ComputingConley}).
          These landmarks will be the vertices of a simplicial
          complex.

	\item Choose a value for the $\eps$ parameter that
          satisfies the requirement \Eq{AlphaBound} for
          \Thm{WitnessSubsetAlpha} and use witness/landmark
          relationships to simultaneously define a simplicial complex
          $\cW_\eps(\Gamma, L)$ and a simplicial multivalued
          map, \cFW. Note that, even though we never
          need to construct an $\alpha$-complex, \Eq{AlphaBound}
          implies that there is an $\alpha$ that
          has the same homology as $\cW$.

	\item Pick a subset of $L$ as a starting guess for an isolated
          invariant set and use Algs.~\ref{alg:invariant:part},
          \ref{alg:grow:isolating}, and \ref{alg:IndexPair}
          of \App{ComputingConley} to find an index
          pair $(|N|,|E|)$ for $f$.  There are many different
          strategies for choosing the initial guess; if one is
          attempting to find periodic orbits, for instance, it makes
          sense to search for recurrent points in the time series and
          use the nearest landmark as the starting point for the
          algorithms.  The important property is that the guess should
          be a subset of its period-image under the simplicial multivalued
          map.

	\item Use a chain selector for \cFW to calculate
          $f_*:H_*(|N|,|E|) \to H_*(|N|,|E|)$.

\end{enumerate}

\noindent In the rest of this section, we illustrate this procedure on H\'enon's
classic map \cite{Henon76}:

\beq{HenonMap}
	 f(x, y) = (y + 1 - 1.4x^2, 0.3x).
\eeq

This map has an invariant set $\Lambda$ that is an
attractor, and we generate a trajectory $\Gamma$ of length $T= 10^5$
that starts from the initial condition $z_0 = (-0.4, 0.3)$, near
$\Lambda$.  The trajectory is shown in \Fig{HenonTraj}. As a simple
test of the witness map technique presented in the previous section,
we use this trajectory to verify the trivial fact that $f$ has a fixed
point. We will assume that $\Gamma$ is an exact
trajectory of \Eq{HenonMap}, making no claim that our computation is
rigorous. The latter could, at least in principle, be done using
interval arithmetic.  Given \Eq{HenonMap}, of course, a simple
calculation shows that this system has \emph{two} fixed points.  Our
goal is to find one of those fixed points using only the time series
$\Gamma$.  Note that this is a proof-of-concept example, not an
exhaustive exploration of the parameter space of the algorithm.
Moreover, it is simple enough that the homology calculations can be
carried out by hand.

\InsertFig{HenonTraj}{Time-series data, $\Gamma$, comprising $10^5$ iterates of the
  H\'enon map \Eq{HenonMap} (grey points) from an initial
  condition near the attractor, and the set $L$ of $216$ evenly
  spaced landmarks (blue circles) that approximate the attractor from
  which $\Gamma$ was sampled. Note that the vertical and horizontal
  scales are different.}{HenonTraj}{4in}

We begin by selecting a set of landmarks to approximate the attractor $\Lambda$.
Again, as a proof of principle, we simply space these landmarks evenly
within the bounding box of the orbit $[-1.5,1.5] \times [-0.4, 0.4]$.
With the goal of reflecting the structure of the attractor and yet
having significantly fewer landmarks than points on the orbit ($\ell
\ll T$), we use a hexagonal grid with spacing $\beta = 0.05$. Indeed,
retaining only those landmarks that are within $\beta$ of a
time-series point gives $\ell = 216$ landmarks (so $\ell \sim
\sqrt{T}$), as shown in \Fig{HenonTraj}. This has the effect of
distributing the landmarks across the attractor with enough resolution
to detect some of its fractal structure.

The next step in the process is to define witness-landmark relation
$R_\eps$ of equation \Eq{FuzzyWitness}.
Although we do not need to build an $\alpha$-complex, it is
possible to do so when the requirement \Eq{AlphaBound} is satisfied. For example,
given the hexagonal geometry, the
$\alpha$-complex will be a clique complex when $\alpha <
\beta/2$---i.e., when it is trivially totally disconnected---or when $\alpha
\ge \beta/\sqrt{3}$, the distance from a vertex to the center of the
equilateral triangle of side $\beta$. Similarly, by construction,
every data point is in one of the equilateral triangles formed from
the landmarks, i.e., $M \le \frac{\beta}{\sqrt{3}}$. The requirement
\Eq{AlphaBound} is then satisfied if we choose $\alpha = \beta/\sqrt{2}$, and

\[
	\eps \le \left(\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{3}} \right) \beta.
\]

Recall that given an $\eps \ge 0$, a pair $(x_t,l_j) \in R_\eps \subset \Gamma \times L$ ($x_t \in W_\eps(\Gamma,l_j)$), according to \Eq{FuzzyWitness}, if and only if $d(x_t, l_j) \le d(x_t,L) + \eps$.
This witness relationship serves to define a
clique complex $\cW_\eps(\Gamma,L)$, by \Eq{CliqueComplex}: the edge $\langle
l_i,l_j\rangle \in \cW_\eps$ if and only if $W_\eps(\Gamma, l_i) \cap W_\eps(\Gamma,l_j) \neq
\emptyset$. By varying $\eps$ up to the bound above and looking at the resulting
complexes, we finally selected $\eps = 0.005$; this is large enough so that the complex is connected, but small enough that the shape of the complex still reflects the primary fold in the attractor. The resulting complex is shown in \Fig{HenonFixedPt}.

Having built the witness complex, we then construct the cellular
multivalued map $\FW: |\cA_\alpha(L)| \rightrightarrows |\cA_\alpha(L)|$ using the
witness-landmark relationships, as described by \Eq{CellularWitnessMap}.  Next
we use the time series to search for an isolating neighborhood for
$\FW$. To apply \Alg{grow:isolating}, we need a guess for an isolating
neighborhood.  For a periodic orbit, this can be found by looking for
nearly recurrent points in the time series
\cite{Lathrop1989}. Consequently, to choose an initial guess for the
purpose of finding a fixed point, we can simply search for a time $t$
that minimizes $d(x_t, x_{t+1})$. Following this approach, we find
that $x_{39,436} = (0.6313, 0.1894)$ is a good candidate---indeed, it
is close to the analytical fixed point $(x^*,y^*) \approx
(0.6313544771, 0.1894063431)$.
The cell of the landmark nearest to this point gives a useful initial
guess for the isolating neighborhood for \Alg{grow:isolating}.
This isolating neighborhood is then
used as the input of \Alg{IndexPair} to obtain an index pair $(N,E)$
for the fixed point of $f$. The result is shown in \Fig{HenonFixedPt}.

\InsertFig{HenonFixedPt}{Witness complex and an index pair $(N,E)$ for
  the fixed point of the H\'enon map. The points correspond to the
  landmarks, $L$, the blue squares represent the isolating
  neighborhood $N \setminus E$ and the red triangles are the
  exit set $E$.  The witness complex corresponds to the landmarks, the
  lines, and the grey triangles.}  {HenonFixedPt}{4in}

The final step is to calculate the Conley index of the isolated
invariant set that is the invariant part of $N \setminus E$.  From
this index, we can then infer the existence of a fixed point.  We
begin by taking a close look at the map \cFW restricted to the index
pair $(N,E)$. The index pair is shown in \Fig{HenonFixedPt}. Recall
that $\FW$ is a map that is constant on $\alpha$-cells. Though we do not
need to compute these $\alpha$-cells, visualizing them---as in the
sketch shown in \Fig{IndexPairGrid}---helps in understanding the
various multivalued maps involved in this process.  In
\Fig{IndexPairGrid}, $N = \{A_1,\ldots,A_9\}$ and $E =
\{A_1,A_2,A_7,A_8\}$. The blue and red landmarks are the nexuses of
the $\alpha$-cells that make up $N\setminus E$ and $E$, respectively.

In this example, the map $\FW$ restricted to the
index pair can be described by the following transition matrix:

\beq{TransitionMatrix} S = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 &
  0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1
  & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 &
  1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0
  & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &
  0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
		\end{pmatrix},
\eeq

where $S_{ij} = 1$ if and only if $A_j \subset
\FW(\Int(A_i))$---i.e., there is a witness of the landmark associated
with $A_i$ whose image under the shift map is a witness of the
landmark associated with $A_j$. For example $\FW(A_6) =
A_2\cup A_3\cup A_8\cup A_9$.
Geometrically, this image is a disk, and is thus acyclic. Indeed, it
is straightforward---if tedious---to verify from \Eq{TransitionMatrix}
that each cell maps to an acyclic set under the witness map
$\FW|_N$. Moreover, by \Eq{CellularWitnessMap}, whenever $x \in A_i \cap A_j$,
it has an image that is the intersection of the images of the
individual cells. In this way, one can compute the images under $\cFW$
of the twelve one-dimensional simplices in the index pair. These
images, inferred from \Eq{TransitionMatrix}, can be verified to be
acyclic when restricted to $N$. Finally only one of the two-simplices
in $N$ remains in $N$ under the map, namely
$\cF_W(\simplex{l_3,l_4,l_9}) = \simplex{l_6,l_7}$; this image has no
homology.  Consequently, the map $\FW$ is acyclic \emph{on the index
  pair}. This condition is sufficient for our limited purpose here of
verifying that $f$ has a fixed point in $N$.  More generally, the
acyclicity of the witness maps on the entire complex could easily be
automated---as is done for cubical complexes in the software package
``CHomP" \cite{CHOMP}.

\InsertFig{IndexPairGrid}{The witness complex (grey triangles, black
  lines, and dots) for the index pair $(N,E)$ for the landmarks from
  \Fig{HenonFixedPt}.  In this case, the witness complex is also the
  nerve of the $\alpha$-cells (shown as truncated red and blue spheres
  in the figure) since \Eq{AlphaBound} is satisfied.  Dashed lines
  show the analytical image of the simplicial complex under the
  H\'enon map \Eq{HenonMap}. The black square is the fixed point.}
          {IndexPairGrid}{5.5in}

Now, we want to represent this index pair with a corresponding
simplicial complex $\cW(N,E)$---specifically, the witness complex
associated with the landmarks $\{l_1,\ldots,l_9\}$, corresponding to
$\alpha$-cells $\{A_1,\ldots,A_9\}$. The witness complex
$\cW(N,E)$, which is also the $\alpha$-complex in this case,
is pictured in \Fig{IndexPairGrid}.
In order to compute the Conley index, we need a simplicial complex
that represents the quotient space $N/E$. Since the $\alpha$-cells
$A_1, A_2, A_7$, and $A_8$ make up the exit set $E$, we make the
identification
\[
 	l_1 \cong l_7 \cong l_2 \cong l_8 := E.
\]
The resulting simplicial complex is shown in \Fig{QuotientComplex}.
The homology of the simplicial complex $\cK(N,E)$ can easily
be computed by hand in this example. In particular, the quotient space
$N/E$ consists of a single connected component so $H_{*0}(N,E) =
\bZ_2$. The quotient space has a single, nonbounding cycle:
\[
	\sigma = \simplex{E,l_3} + \simplex{l_3,l_4} + \simplex{l_4,l_5} +
	\simplex{l_5,l_6} + \simplex{l_6,E},
\]
so $H_{*1}(N,E) = \bZ_2$.

\InsertFig{QuotientComplex}{The quotient simplicial complex
  $N\setminus E$, where $N$ is the simplicial complex shown in
  \Fig{IndexPairGrid} and $E= \{l_1,l_2,l_7,l_8\}$.}{QuotientComplex}{2.0in}

We are now ready to compute the Conley index of the isolating
neighborhood $N \setminus E$. Specifically, we compute $f_*: H_*(N,E)
\to H_*(N,E)$. As described in the previous section, the homology of
$f$ is equivalent to the homology of $\varphi$, where $\varphi$ is a
chain selector for \cFW. Therefore, in order to compute the Conley
index, we need to find $\varphi_*([\sigma]) := [\varphi(\sigma)]$.

The chain selector $\varphi$ is defined inductively by first
determining the image of each vertex in $\cK(N,E)$ (to be
enclosed by \cFW), and then determining the image of each edge so that
$\varphi$ commutes with the boundary operator. In addition, recall
that \cFW on the quotient space must be an enclosure of the index map
$f_N$. We begin with the initial assignment of vertices:
\begin{align*}
	\begin{array}{l |cccccc}
	\simplex{i} & \simplex{E} & \simplex{l_3} & \simplex{l_4} & \simplex{l_5}
			& \simplex{l_6} & \simplex{l_9} \\
	\hline
	\varphi_0(\simplex{l_i}) & \simplex{E} & \simplex{l_6} & \simplex{l_5} & \simplex{l_3}
			& \simplex{l_9} & \simplex{l_6}
	\end{array}
\end{align*}
To compute $\varphi(\sigma)$, we need to find the images of the edges in $\sigma$ as well.

In order for $\varphi$ to be a chain selector for \cFW, the image of each edge,
$\tau$, must be a subset of $\cFW(\tau)$. Furthermore, $\varphi$ must
commute with the boundary operator, so we need $\varphi_0 \circ
\partial_1 = \partial_1 \circ \varphi_1$. Those two conditions yield
the following edge assignments:
\begin{align*}
	\begin{array}{l|ccccc}
		\tau & \simplex{E,l_3} & \simplex{l_3,l_4} & \simplex{l_4,l_5}
			& \simplex{l_5,l_6} & \simplex{l_6,E}\\
		\hline
		\varphi_1(\tau) & \simplex{E,l_6}& \simplex{l_6,l_5}& \simplex{l_5,l_4}+\simplex{l_4,l_3}
			&\simplex{l_3,l_9} & \simplex{l_9,E}
	\end{array}
\end{align*}
It follows that $\varphi(\sigma) = \sigma$ (in $\bZ_2$
homology). Since this map is not nilpotent, then it is not in the
shift equivalence class of $[0]$ and thus the invariant set $S =
\inv(N \setminus E,f) \neq \emptyset$ \cite[Thm 10.91]{KMM04}.

With this very simple example, we have illustrated that the witness
complex and the associated witness map can be used to compute the
Conley index for a simple of isolating neighborhood. We plan in
the future to apply this method to more-complex and higher-dimensional
dynamics.

config: arxiv


TITLE: Fermionic Poisson bracket
QUESTION [3 upvotes]: I'd like to understand the Poisson bracket for fermions in classical field theory defined on a cylinder (with coordinates $(t,x)$, $x$ being the compact direction) and propagating on $T^n$ with constant metric $G_{ij}$ (though $T^n$ probably isn't important to the discussion here).
The action is $$S = \frac{i}{4\pi}\int dt dx \,\,G_{ij} \left[ \psi^i(\partial_t + \partial_x)\psi^j + \bar{\psi}^i(\partial_t - \partial_x)\bar{\psi}^j\right] .$$
From this how can I give a general definition of the Poisson bracket? Some constraints on it should be the following:

It should be symmetric (since in the bosonic case it's anti-symmetric).
I should be able to recover the standard relation $$\{\psi^i(t,x), \psi^j(t,x') \}_{PB} = -2\pi i G^{ij}\, \delta(x-x')$$ from it.

One definition which seems to work is  $$\{F,G\} = -2\pi i \int dx\,\, G^{ij}\Big(\frac{\delta F }{\delta \psi^i } \frac{\delta G}{\delta \psi^j} +  \frac{\delta F }{\delta \bar{\psi}^i } \frac{\delta G}{\delta \bar{\psi}^j}\Big)$$
However, if this is right, I'd like to see why on more general grounds.
Also one particular Poisson bracket I'm interested in computing is $\{(\partial_t - \partial_{x_1})\psi^i(x_1), \psi^j(x_2)\}$ which I get from the above definition to be $2\pi i \frac{\partial}{\partial x_1} \delta(x_1 - x_2)$. Is this sensible?
Edit: for reference, I'm looking at Appendix A in the following paper by Kapustin and Orlov:
http://arxiv.org/abs/hep-th/0010293
and trying to verify the Poisson Brackets in eqn. (48).

REPLY [3 votes]: The super-Poisson bracket follows from a super-version of the Dirac-Bergmann or the Faddeev-Jackiw procedure. Diligent care must be taken to achieve consistent sign conventions when dealing with Grassmann-odd variables, see e.g. my Phys.SE answer here.
The singular Legendre transformation for fermions is also discussed in my Phys.SE answer here. In OP's case [which is the fermionic part of eq. (44) in Ref.1], the Lagrangian density reads
$$\tag{1} {\cal L}~=~ \frac{i}{4\pi} G_{ij}\left[\psi^i  \dot{\psi}^j + \bar{\psi}^i  \dot{\bar{\psi}}^j\right]-{\cal H}, $$
where we have identified the Hamiltonian density
$$\tag{2} {\cal H}~=~ \frac{i}{4\pi} G_{ij}\left[\psi^i \partial_x\psi^j
- \bar{\psi}^i  \partial_x\bar{\psi}^j\right].$$
The symplectic one-form potential can be transcribed from the kinetic term in (1):
$$ \tag{3} \vartheta(t) ~=~\frac{i}{4\pi}\int\! dx~G_{ij}\left[
\psi^i(x,t)~\mathrm{d}\psi^j(x,t)
+ \bar{\psi}^i(x,t) ~\mathrm{d}\bar{\psi}^j(x,t)\right], $$
where $\mathrm{d}$ denotes the exterior derivative in infinitely many dimensions.  The symplectic two-form is then
$$  \omega(t)~=~\mathrm{d}\vartheta(t) ~=~\frac{i}{4\pi}\int\! dx~G_{ij}\left[ \mathrm{d}\psi^i(x,t)\wedge\mathrm{d}\psi^j(x,t) +\mathrm{d}\bar{\psi}^i(x,t) \wedge\mathrm{d}\bar{\psi}^j(x,t)\right]$$
$$ \tag{4} ~=~\frac{i}{4\pi}\int\! dx~dy~ G_{ij}~ \delta(x-y) \left[
\mathrm{d}\psi^i(x,t)\wedge\mathrm{d}\psi^j(x,t)+\mathrm{d}\bar{\psi}^i(x,t)\wedge\mathrm{d}\bar{\psi}^j(x,t)\right].\qquad $$
The equal-time super-Poisson/Dirac bracket on fundamental fields is the inverse supermatrix of the supermatrix for the symplectic two-form (4):
$$ \tag{5} \{\psi^i(x,t), \psi^j(y,t)\}_{PB}~=~ \frac{2\pi}{i}(G^{-1})^{ij} ~\delta(x-y)~=~\{\bar{\psi}^i(x,t), \bar{\psi}^j(y,t)\}_{PB}, $$
and other fundamental super-Poisson brackets vanish. The equal-time super-Poisson bracket read
$$ \tag{6}\{F(t),G(t)\} ~=~ \frac{2\pi}{i}\int \! dx  (G^{-1})^{ij}\left[\frac{\delta_R F(t)}{\delta \psi^i(x,t)} \frac{\delta_L G(t)}{\delta \psi^j(x,t)}+\frac{\delta_R F(t)}{\delta \bar{\psi}^i(x,t)} \frac{\delta_L G(t)}{\delta \bar{\psi}^j(x,t)} \right]\qquad $$
for arbitrary equal-time functionals $F(t)$ and $G(t)$. Here the subscripts $L$ and $R$ refer to differentiation from left and right, respectively.
References:

A. Kapustin and D. Orlov, arXiv:hep-th/0010293.


TITLE: Proof that covariant derivative of contravariant components of metric vanish for metric compatabilit
QUESTION [0 upvotes]: In this excercise I want to show that $\nabla_\rho g_{\mu \nu}=0$ and $\nabla_\rho g^{\mu \nu}=0$
This should probably be very easy, but excuse me I'm completly new to GR.
So to do this I used that the covariant derivative of a rank 2 tensor is
$$
\nabla_c t^{ab} = \partial_c t^{ab} + \Gamma^a_{dc}t^{db}+\Gamma^b_{dc}t^{ad}
$$
$$
\nabla_c t_{ab} = \partial_c t_{ab} - \Gamma^d_{ac}t_{db}-\Gamma^d_{bc}t_{ad}
$$
And I rewrote the connection in therms of partial derivatives of the metric.
So for the covariant components of the metric this worked out nicely, everything canceld and ended up to be zero.
Now with the contravariant components I'm running into trouble:
$$
\nabla_\rho g^{\mu \nu} = \partial_\rho g^{\mu\nu} + \Gamma^\nu_{d\rho}g^{d\nu}+\Gamma^v_{d\rho}g^{\mu d}=\\ \partial_\rho g^{\mu\nu} + \frac{1}{2}g^{\mu e} (\partial_d g_{e \rho} +
\partial_\rho g_{de} -\partial_e g_{d\rho})g^{d \nu}+
\frac{1}{2}g^{\nu e} (\partial_d g_{e \rho} + \partial_\rho g_{de} -\partial_e g_{d\rho})g^{\mu d}
$$
Now the two parts inside the brakets are the same and the metrics outside replace the dummy indices d and e by $\nu$ and $\mu$. If I raie the indices d and e four of the six terms cancel each other but then I am left with:
$$
\nabla_\rho g^{\mu \nu} = 2 \partial_\rho g^{\mu \nu}
$$
This is clearly not zero for a general metric in non local coordinates. So I assume I'm messing something up completly, pleas ehelp me to understand this.

REPLY [2 votes]: Actually you were close but no cigar.
The purpose of this post to introduce you to the dual space connections.
In addition, I don't agree with the examples of covariant derivative of tensors which you provided in your question. For instance, taking the covariant derivative of a $(r,s)$ tensor changes it to a $(r,s+1)$ tensor.
I'm going to use $D$ to denote the covariant derivative (fewer keystrokes) - and I'm assuming the Levi-Civita connection (i.e., the connection is torsion free.)
The connections in the dual space are denoted as $\omega^{j}_{i}=\Gamma^{j}_{ik} dx^{k}.$
And I'm using the full metric tensor
$$g=g_{ij}dx^{i}\otimes dx^{j}$$
which are typically denoted by the tensor coefficients only, for example, $g_{ij}.$
The covariant derivative of the metric tensor
$$Dg=0$$
ensures the magnitudes of the tangent vectors and the angles between them remain invariant when the metric tensor is parallel transported.
Expanding the calculation of the covariant derivative
$$
Dg =D(g_{ij})dx^{i}\otimes dx^{j}+g_{ij}D(dx^{i})\otimes dx^{j}+g_{ij}dx^{i} \otimes D(dx^{j})
$$
yields
$$
Dg=(dg_{ij}-\omega^{k}_{i}g_{kj}-\omega^{k}_{j}g_{ik})\otimes dx^{i}\otimes dx^{j}=0
$$
which implies
\begin{align}
dg_{ij}=&\; \omega^{k}_{i}g_{kj}+\omega^{k}_{j}g_{ik}\\
\Gamma^{j}_{ik}=&\; \Gamma^{j}_{ki}
\end{align}
where
\begin{align}
D(g_{ij})\;=&\;dg_{ij}\\
D(dx^{i})=&-\omega^{k}_{i}g_{kj}dx^{i}\\
D(dx^{j})=&-\omega^{k}_{j}g_{ik}dx^{j}.
\end{align}
At this point, I should probably also prove that when $2$ vector fields from the tangent space are parallel transported along a curve with respect to the metric then $DG=0$ - but I don't have time today - and it's a seperate question.
Also, it's possible to derive the Christoffel symbols from equation $(1)$ if one writes $dg_{ij}$ in terms of partials, the $\omega^{i}_{j}$ connections in the form $\Gamma_{ijk}$ and use the metric to raise the $k$ at the very end. But it can be a little tricky.
Ref: S.S. Chern, W.H. Chen, and K.S. Lam, $Lectures\;on\;Differential\;Geometry$, Series On University Mathematics, Vol. $\mathbf{1}$ (1998), 133-139.