Christian Marks

Ecumenical dispatches from the London Library

Archive for the ‘Bagatelle’ Category

A co-topological category

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This post describes a nearly useless category that reminds me of the topologies associated with categories of coalgebras for a set endofunctor.

Let {X, Y} be topological spaces. A not-necessarily continuous map {f:X\rightarrow Y} is open if for every open set {U\subset X}, {f[U]} is open in {Y}. The category {\mathbf{Opt}} of topological spaces and open maps has the same isomorphism types as the category {\mathbf{Top}} of topological spaces and continuous maps.

Proof: A map is an isomorphism in {\mathbf{Opt}} if and only if it is a homeomorphism in {\mathbf{Top}}. \Box

Coproducts exist in {\mathbf{Opt}}. In general, products do not exist in {\mathbf{Opt}}. However, something analogous to powers of a single space exist.

For a topological space {X}, {\square X} (“the square of {X}“) denotes the topological space with underlying set {X\times X} and topology generated by the basis

\displaystyle \{ U\times V, \Delta_X[U] : U,V\,\mathrm{open}\,\mathrm{in}\,X\}

where {\Delta_X:X\rightarrow X\times X} is the diagonal. This topology is the smallest topology on {X\times X} making the diagonal map open (and continuous).

The {\square} construction defines an endofunctor {\square} on {\mathbf{Opt}} together with a natural transformation {\delta_X:X\rightarrow\square X} (the same underlying map as {\Delta_X}) from the identity functor {1_\mathbf{Opt}} to {\square}. There is a one-to-one correspondence between maps {f:X\rightarrow Y} and maps {g:X\rightarrow\square Y} such that {\pi_1 g= \pi_2 g} for {i=1,2}, where {\pi_i:\square Y\rightarrow Y} is the projection (projections are open).

Proof: Given a open map {f:X \rightarrow Y}, the open map {\delta_Y f} satisfies {\pi_1\delta_Y f = f = \pi_2 \delta_Y f}. Conversely, a map {g:X\rightarrow \square Y} with {\pi_1 g= \pi_2 g} for {i=1,2}, defines {f=\pi_1 g}. Then

\displaystyle \delta_Y f = (\pi_1 g \times \pi_1 g) \delta_Y = (\pi_1 g \times \pi_2 g) \delta_Y = (\pi_1 \times \pi_2) \delta_Y g = g

\Box

Written by Christian Marks

July 6, 2011 at 3:37 AM

Posted in Bagatelle

The tragedy of asymptotic density

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If A\subseteq\mathbb{N}, we define A(n) = A\cap[1,n] and say that A has asymptotic  density if \lim_{n\rightarrow\infty} |A(n)|/{n} exists.  The collection of sets that have asymptotic density does not form an algebra.

The set A = \{4,6, 9, 11, 13, 15, 16, 18, 20, 22, 24, 26, 28, 30, 33,\ldots\} has asymptotic density 1/2. This is illustrated by the blue graph below. The red graph is the intersection of A with the set 2\mathbb{N} of even numbers. Call this intersection B. The upper density of B is \limsup_{n\rightarrow\infty} |B(n)|/{n}=1/3 and the lower density of B is \liminf_{n\rightarrow\infty} |B(n)|/{n}=1/6. (This can be established with certain geometric series.) The two sets A and 2\mathbb{N} have asymptotic density (in this case, 1/2), but their intersection B=A\cap2\mathbb{N} does not.

4 6 9 11 13 15 16 18 20 22 ... 30 33 35 ... 61 63 64 ...

This implies that asymptotic density cannot be used as a probability measure.

Written by Christian Marks

May 31, 2010 at 1:17 AM

Posted in Bagatelle

Spivak’s mistaken problem

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In “Calculus on Manifolds,” Michael Spivak incorrectly states conditions for a linear operator on a finite dimensional real inner product space to be angle preserving, on the assumption that the space has a basis consisting of eigenvectors of the operator.

Let T be an injective linear operator on a finite dimensional real inner product space V. The operator T is angle preserving if and only if for all nonzero x,y\in V,

\frac {\langle T x | T y\rangle} {|T x| |T y|} = \frac {\langle x | y\rangle} { |x| |y| }

Spivak writes that assuming there exists a basis x_1,\ldots,x_n of V and real numbers \lambda_i,\, 1\le i\le n such that T x_i = \lambda x_i, then T angle preserving if and only of the \lambda_i all have the same absolute value.

This is incorrect: the eigenbasis must be orthogonal. One can produce linear operators on \mathbb{R}^2 with real eigenvalues \pm 1 but which do not preserve angles. An example is an operator with the following matrix in the standard basis:

\left(\begin{array}{cc}-1 & 0\\-2 & 1\end{array}\right).

An eignenbasis for this operator is

\left\{\left(\begin{array}{c}0\\1\end{array}\right),\,\left(\begin{array}{c}1\\1\end{array}\right)\right\}.

We can use the eigenbasis to diagonalize the matrix:

\left(\begin{array}{cc}-1 & 0\\-2 & 1\end{array}\right)=\left(\begin{array}{cc}0 & 1\\ 1 & 1\end{array}\right)\left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right)\left(\begin{array}{cc}-1 & 1\\ 1 & 0\end{array}\right).

The possibility that there may be some other orthogonal eigenbasis that we have overlooked is ruled out by the spectral theorem for linear operators. The spectral theorem for a linear operator T on a real finite dimensional inner product space V states that V has an orthonormal basis consisting of eigenvectors of T if and only if T is self-adjoint. In the example above, the matrix is not symmetric, so there is no orthogonal basis of \mathbb{R}^2 consisting of eigenvectors for our matrix.

Let’s prove that T angle preserving implies that all eigenvalues have the same absolute value. If x and y are two eigenvectors of T with distinct eigenvalues a and b, then there are two cases to consider.

Case one: \langle x | y\rangle \ne 0. We may assume that x and y have norm 1. Apply the Gram-Schmidt procedure to x and y (up to scale). We have that

\langle x | y - \langle x| y\rangle x\rangle=\langle x | y\rangle - \langle x | y\rangle \langle x | x\rangle = 0.

Since T is angle preserving, it follows that

0=\left\langle T x | T (y - \langle x| y\rangle x)\right\rangle = a b\langle x| y\rangle - a^2\langle x| y\rangle.

Dividing by a \langle x | y\rangle, which is nonzero (T is injective and cannot have 0 as an eigenvalue, and we are in case one), we have that a = b.

Case two: \langle x | y\rangle = 0. In that case \langle x + y| x-y\rangle = 0, and since T is angle preserving, \langle T(x + y)|T( x-y)\rangle = 0. This implies that a^2 = b^2, so that a and b have the same absolute value.

Hence in either case, the eigenvalues of T satisfy \lambda_i=\pm\lambda. We may write

V = \ker (T-\lambda I)\oplus\ker (T+\lambda I)

where \ker (T+\lambda I)=\ker (T-\lambda I)^\perp, and apply the Gram-Schmidt orthonormalization process to bases for each summand  to produce an orthornormal basis of V consisting of eigenvectors for T.

The converse, on the assumption that T has an orthonormal eigenbasis, is a straightforward calculation.

Written by Christian Marks

July 6, 2009 at 4:38 AM

Posted in Bagatelle

Skew symmetric matrices

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A skew symmetric matrix of odd order vanishes. Suppose that A is a skew symmetric matrix of even order. If B is the matrix obtained from A by adding the same number \lambda  each of the entries of A, then A and B have the same determinant. The determinant of A equals the following.

\left|\begin{array}{cccc}1&0&\cdots&0\\ 0\\ \vdots&&A\\ 0 \end{array}\right|=\left|\begin{array}{cccc}1&\lambda&\cdots&\lambda\\ 0 \\ \vdots&&A\\ 0 \end{array}\right|=\left|\begin{array}{cccc}1&\lambda&\cdots&\lambda\\1\\ \vdots&&A+\Lambda\\1\end{array}\right|.

where \Lambda is the n\times n matrix with all entries equal to \lambda. The last determinant equals the following sum, using bilinearity.
\left|\begin{array}{cccc}0&\lambda & \cdots&\lambda\\1\\ \vdots&&A+\Lambda\\1\end{array}\right|+\left|\begin{array}{cccc}1&\lambda&\cdots&\lambda\\ 0\\ \vdots & & A+\Lambda \\ 0 \end{array}\right|

Add -\lambda times first column of the first term to the remaining columns, and multiply the first column (hence the determinant) by -\lambda. The result is a skew symmetric matrix, which must be zero since the determinant of a skew symmetric matrix of odd order is zero.
\left|\begin{array}{cccc}0& \lambda & \cdots & \lambda\\ -\lambda\\ \vdots & &A \\ -\lambda \end{array}\right|+\left|\begin{array}{cccc}1&0&\cdots&0\\ 0\\ \vdots & & A+\Lambda \\ 0 \end{array}\right|=|A+\Lambda|

Written by Christian Marks

July 5, 2009 at 7:23 AM

Posted in Bagatelle

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