MS 2001: Week 3

October 5, 2011 in MS 2001 | Leave a comment

I am emailing a link of this to everyone on the class list every Wednesday morning. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

We covered from Proposition 1.4.3 to Proposition 2.1.5 inclusive. Here are the proofs of the product and quotient  rules for the calculus of limits.

Problems

You need to do exercises – all of the following you should be able to attempt. Do as many as you can/ want in the following order of most beneficial:

Wills’ Exercise Sheets

Q.8, 9 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise1.pdf

More exercise sheets

Q. 4 from Problems

Past Exam Papers

Q. 1(a), 2 from http://booleweb.ucc.ie/ExamPapers/exams2010/MathsStds/MS2001Sum2010.pdf

Q . 1(a), 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/exams2009/MathsStds/MS2001s09.pdf

Q. 1(a), 2 fromhttp://booleweb.ucc.ie/ExamPapers/exams2009/MathsStds/Autumn/MS2001A09.pdf

Q. 1(a) fromhttp://booleweb.ucc.ie/ExamPapers/exams2008/Maths_Stds/MS2001Sum08.pdf

Q. 1(a), 2(a)(i), (b) fromhttp://booleweb.ucc.ie/ExamPapers/Exams2008/MathsStds/MS2001a08.pdf

Q. 2 fromhttp://booleweb.ucc.ie/ExamPapers/exams2007/Maths_Stds/MS2001Sum2007.pdf

Q. 1, 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/MS2001Sum06.pdf

Q. 1, 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/Autumn/ms2001Aut.pdf

Q. 1, 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001.pdf

Q. 1, 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001Aut05.pdf

Q. 1(a), 2(a), 3(b) fromhttp://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/MS2001aut.pdf

Q. 1(a), 2 fromhttp://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 1(a), 2(a), 3 fromhttp://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/MS2001.pdf

Q. 1(a), 3 fromhttp://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/ms2001aut.pdf

Q. 1, 2(a) fromhttp://booleweb.ucc.ie/ExamPapers/exams2002/Maths_Stds/ms2001.pdf

Q. 1(a), 2(a), 3 fromhttp://booleweb.ucc.ie/ExamPapers/exams2001/Maths_studies/MS2001Summer01.pdf

Q. 1(a), 2 fromhttp://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf

Tutorial Quiz Questions

These are not necessarily of exam standard but are more an exercise to help your understanding. The quiz we never did this week as ye actually did some exercises and asked questions.

MS 2011: Week 2

September 28, 2011 in MS 2001 | 1 comment

We covered from Proposition 1.1.4 about the inequality relation — up to but not including Proposition 1.4.3.

I will make a decision about tutorial clashes on or about Monday.

Exercises

Q. 1,2 and 4 – 7 from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise1.pdf

Q. 2 – 3 from Problems

An alternative quantisation of a Markov chain? or Why do we need Coalgebras?

September 21, 2011 in Random Walks on Finite Quantum Groups, Research | Leave a comment

Quantisation

I had been of the understanding that a quantisation looks as follows. There is some process or property P(X) of a space X which we want to examine. Depending on the type of space X, from a suitable algebra of complex functions on X, F(X), we can recover and examine many of the properties of P(X) by instead looking at F(X): we essentially have the identification X\leftrightarrow F(X). When the process/ property P(X) is about a space then it is said to be classical or commutative because for any x\in X and f,\,g\in F(X) we have that fg=gf because fg(x)=f(x)g(x)=gf(x) as the f(x),\,g(x) lie in the commutative algebra \mathbb{C}.

Now from X we know about F(X) and vice versa. Now a suitably chosen F(X) is just a commutative C*-algebra so what about a non-commutative C*-algebra F — can we examine it’s “underlying space” in the same way?

So essentially, this means that I thought you quantised objects, such as Markov chains, by replacing a commutative C*-algebra F(X) with a not-necessarily commutative one. (This roughly follows my interpretation as per this)

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MS 2001: Week 1

September 21, 2011 in MS 2001 | Leave a comment

If you have clashes with the Tuesday tutorial starting on 4/10/11 please contact me ASAP.

Some useful links:

Stephen Wills MS 2001 homepage.

Stephen Wills: Supplementary Notes.

Past Exam Papers.

This week’s exercises

Q.3 from Exercise Sheet 1.

MS 2001: Lecture Notes

September 20, 2011 in MS 2001 | Leave a comment

The lecture notes are available from An Scoláire, College Road (across from the main College Road entrance to UCC). They are for sale at €12 — just ask for the notes for MS2001.

http://www.goldenpages.ie/an-scolaire-cork-city/2/

 

The Mathematics of Card Shuffling

September 19, 2011 in General | Leave a comment

I’m giving a talk in Blackrock Castle Observatory on October 7. See the link for more details.

http://www.bco.ie/2011/09/first-fridays-at-the-castle-celebrating-50-years-of-human-spaceflight/

Random Walks on Finite Quantum Groups: Random Walks on Comodule Algebras

September 15, 2011 in Random Walks on Finite Quantum Groups, Research | 1 comment

Taken from Franz & Gohm.

Let return to the map b:M\times G\rightarrow M considered in the beginning of the previous section. If G is a group, then b:M\times G\rightarrow M is called a left action of G on M, if it satisfies the following axioms expressing associativity and unit (\checkmark0),

b(b(x,g),h)=b(x,hg), and b(x,e)=x

for all x\in M, g,h\in G; and where e\in G is the identity. As before we have the unital *-homomorphisms \alpha_g:F(M)\rightarrow F(M). Actually, in order to get a representation of G on F(M), i.e. \alpha_g\alpha_h=\alpha_{gh} for all g,\,h\in G we modify  the definition and use \alpha_g(f)(x):=f(b(x,g^{-1})). (Otherwise we get an anti-representation. But this is a minor point at this stage). In the associated coaction \beta:F(M)\rightarrow F(M)\otimes F(G) the axioms above are turned into the coassociativity and counit properties. These make perfect sense not only for groups but also for quantum groups and we state them at once in this more general setting. We are rewarded with a particular interesting class of quantum Markov chains associated to quantum groups which we call random walks and are the subject of this lecture.

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Random Walks on Finite Quantum Groups: Quantum Markov Chains

September 14, 2011 in Random Walks on Finite Quantum Groups, Research | 2 comments

Taken from Random Walks on Finite Quantum Groups by Franz & Gohm.

To motivate the definition of quantum Markov chains let us start with a reformulation of the classical Markov chain. Let M,\,G be finite sets. Any map b:M\times G\rightarrow M may be called an action of G on M. Let F(M) and F(G) be the *-algebra of complex functions on M and G. For all g\in G, we have unital *-homomorphisms (\checkmark) \alpha_g:F(M)\rightarrow F(M) given by (\alpha_g f)(x):=f(b(x,g)). They can be put together into a single unital *-homomorphism

\displaystyle\beta:F(M)\rightarrow F(M)\otimes F(G), \displaystyle f\mapsto \sum_{g\in G}\alpha_g f\otimes\mathbf{1}_{\{g\}},

where \mathbf{1}_{\{g\}} is the indicator function. We get a natural non-commutative generalisation just by allowing the algebras to become non-commutative (by replacing the C*-algebras F(M) and F(G) by more general (!), not necessarily commutative C*-algebras).

Let B and A be unital C*-algebras and \beta:B\rightarrow B\otimes A a unital *-homomorphism. Here B\otimes A is the spatial tensor product. Then we can build up the following iterative scheme for n\geq 0:

j_0:B\rightarrow B, b\mapsto b

j_1:B\rightarrow B\otimes A, b\mapsto\beta(b)=b_{(0)}\otimes a_{(0)}.

(Sweedler’s notation b_{(0)}\otimes b_{(1)} stands for \sum_i b_{0i}\otimes b_{1i} and is very convenient for writing formulas).

\displaystyle j_n:B\rightarrow B\otimes\bigotimes_{1}^nA, j_n=(j_{n-1}\otimes I_A)\circ\beta,

\displaystyle b\mapsto j_{n-1}(b_{(0)})\otimes b_{(1)}\in \left(B\otimes\bigotimes_1^{n-1}A\right)\otimes A.

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Direct Limits and Tensor Products: Tensor Products of C*-algebras

September 9, 2011 in C*-Algebras & Operator Theory, Research | 1 comment

Taken from C*-algebras and Operator Theory by Gerald Murphy.

If H and K are vector spaces, we denote by H\otimes K their algebraic tensor product. This is linearly spanned by the elements x\otimes y (x\in H, y\in K).

One reason why tensor products are useful is that they turn bilinear maps (a bilinear map \varphi has \lambda\varphi(x,y)=\varphi(\lambda x,y)=\varphi(x,\lambda y)) into linear maps (\lambda\varphi(x,y)=\varphi(\lambda x,\lambda y)). More precisely, if \varphi:H\times K\rightarrow L is a bilinear map, where H,\,K and L are vector spaces, then there is a unique linear map \varphi_1:H\otimes K\rightarrow L such that \varphi_1(x\otimes y)=\varphi(x,y) for all x\in H and y\in K.

If \rho,\,\tau are linear functionals on the vector spaces H,\,K respectively, then there is a unique linear functional \rho\otimes\tau on H\otimes K such that

(\rho\otimes\tau)(x\otimes y)=\rho(x)\tau(y)

since the function

H\times K\rightarrow\mathbb{C}, (x,y)\mapsto \rho(x)\tau(y),

is bilinear.

Suppose that the finite sum \sum_jx_j\otimes y_j=0, where x_j\in H and y_j\in K. If y_1,\dots,y_n are linearly independent, then x_1=\cdot=x_n=0. For, in this case, there exist linear functionals \rho_j:K\rightarrow \mathbb{C} such that \rho_j(y_i)=\delta_{ij}. If \rho:H\rightarrow\mathbb{C} is linear, we have

0=(\rho\otimes \rho_j)(\sum_i x_j\otimes y_j)=\sum_i\rho(x_i)\rho_j(y_i)=\rho(x_j).

Thus \rho(x_j)=0 for arbitrary \rho and this shows that all the x_j=0.

Similarly if the finite sum \sum_jx_j\otimes y_j=0 with the x_j linearly independent, implies that all the y_j are zero.

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MS 2001: Selected Limit Questions

August 8, 2011 in MS 2001 | Leave a comment

Questions 1 (ii) — (vii) from here.

With all of these questions the first thing we want to do is plug in the value; i.e. if we are evaluating \lim_{x\rightarrow a}f(x) we should first try f(a). If we find that f(a) is undefined (e.g. division by zero), we will usually have to ‘factor out the bad stuff’. The reason this works is because of the following theorem:

Theorem

Suppose that f(x)=g(x) except perhaps at x=a. Then

\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}g(x).

Take for example

\lim_{x\rightarrow 0}\frac{x^2}{x}.

Now if f(x)=x^2/x, then f(0)=0/0 — which is undefined. However we can cancel an x above and below… well what we really do is as follows:

\frac{x^2}{x}=x\times\underbrace{\frac{x}{x}}_{=1}=x.

However this is only true in the case where x\neq 0; i.e. we have that x^2/x=x for all x\neq 0 — and use the above theorem to say that

\lim_{x\rightarrow 0}\frac{x^2}{x}=\lim_{x\rightarrow 0}x=0.

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