TIFR Bombay – Ph.D. Mathematics Interview

Post the qualifying entrance exam stage for the Ph.D. program in Mathematics at TIFR Bombay, I decided to note down my experiences at the interview that was to follow.

Pre interview stage – I have readied myself for the fate that my chances of getting accepted to this program are minuscule at best and none for the most part at this current stage. Considering the competitive nature of admissions for a Ph.D. at TIFR Bombay, I think this is a pretty accurate view.

I reach on time for the interview, the institute intimidates me a bit with its extremely long passageways and silence which I thought would be an amazing place to do research. I think that TIFR is an amazingly well furnished and well designed place with a small but serene campus with a beach to boot.

I am the first person to be interviewed for the day, my interview panel consisting of five professors: 4 male, 1 female; 2 relatively young and 3 older. I describe the conversation that follows as per my memory. I stands for interviewer; A stands for me.

I:  Good morning. You are doing B.Sc.(III)?
A: Good morning. Yes sir, I am, from St. Xavier`s College, Mumbai.
I: Why did you apply to TIFR?
A: Since I have been in college, I have been hearing about TIFR being one of the best institutes to study mathematics at the graduate level in India. That is why I applied.
I: Why not apply after an M.Sc.?
A: I was under the impression that I could do M.Sc. coursework in TIFR under the Ph.D. program..(?)
I: (Laughs) Many students who come for interviews have these misconceptions, they do not realize that we do not provide an Integrated Ph.D. but rather a direct Ph.D. program.
(Pause) Don`t worry, many students from your college have ended up doing a Ph.D. here after B.Sc.
I: What are you primary interests?
A: Linear Algebra, Metric Spaces, Topology.
I: What topic are you comfortable with?
A: I have no particular preference sir. (conscious of being scrutinized )
I: Okay then, what in Topology have you done?
A: Metric Spaces,  point set topology.
I: Which book have you referred to?
A: I am following Munkres – Topology and Klaus Janich – Topology.

I: Okay, Give me an example of a nonmetrizable topology?
A: The set of Real Numbers under co-finite topology is non metrizable. (A sequence 1/n converges to every real number in the cofinite topology, which no metric on R would allow.)
I: Are and homeomorphic?
A: (Pause) No.. For suppose there existed a homeomorphism then consider a line in and its preimage in . Delete both these sets then one will be disconnected while the other will still be connected.
I: What if the preimage of the line is a plane? (Space filling Curve?)
A: Sir, that would not be possible since it is a homeomorphism, the homeomorphic image of a line would be a bijective path. The space filling curve is not injective. But in general to prove that the set would still be connected after removing the preimage..
I: (Waits) (Says to the other professor – perhaps he hasn`t done it yet?) Have you studied the fundamental group? Your intuition is correct, but this method of proof can be quite difficult to prove.
A: No sir, I have not studied algebraic topology yet.
I: Okay, if you take a point from 2-d euclidean plane and take a circle around it, you cannot shrink it to a point, but you can do so in 3-d plane.
I: Give an example of a continuous bijection whose inverse is not continuous?
A: Consider F:R(discrete) to R  identity map.
I: Can you provide the topological conditions on the domain and range such that a continuous bijection will always be a homeomorphism?
A: If the domain is compact and the Range is Hausdorff.
I: Prove it on the board.
A: (Prove it on the blackboard)
I: Is every compact metric space complete?
A: Yes. (Use sequential compactness as an equivalent condition)
I: Prove sequential compactness then.
A: (Chokes)(Proves with hint).
I: Is it possible to induce a metric topology on (0,1) which is complete?
A: Yes. Completeness is not a topological property..
I: Provide an example.
A: (thinking)
I: Is (0,1) homeomorphic to R?
A: (Oh!) Yes! (Writes down the homeomorphism on the board.)
I: Okay, now how would you answer the previous question?
A: Let F be the homeomorphism from (0,1) to R. Consider the metric on (0,1) : D(x,y) = |F(x)-F(y)|.
I: Good.
I: Is every complete metric space compact?
A: No. Consider R under usual or discrete metric.
I: Okay some analysis now. Let F_n be a sequence of functions C[0,1] converge point wise to F in C[0,1]. Does ?
A:(Could not answer)
I: Okay you can think about it later. A continuous function on R has the set of rationals in its zero set. What can you say about the function?
A: The function is the zero function (because of sequential continuity).
I: Describe a function that is continuous only on the Irrationals.
A:  (thinks) Consider the function which is zero on all the irrational numbers and on the rational numbers it takes the value f(p/q) = 1/q where gcd (p,q) = 1. Thissequence converges only on the irrational numbers.
I: (Nodding) What about 0?
A: 0 can be considered as 0/1, I can assign the value 1 according to my definition, in principle any nonzero value would do.
I: Have you studied Measure theory?
A: No.
I: Fourier Analysis?
A: Very basic Cesaro sum converges to function, weirstrass approximation theorem.
I: So no big convergence theorems.
A: No.
I: A linear transformation which is an Isometry on R3 and orientation preserving is rotation about an axis passing through the origin. Can you prove this?
A: (thinks)
I: Do you know what an isometry is? An isometry means a distance preserving map.
A: Since the transformation is on R3, its characteristic polynomial is of degree 3 and has atleast one real root. Since the map is an isometry it is bijective and hence has a trivial kernel. Thus the real eigenvalue cannot be 0. This would mean that there is a nontrivial invariant line. Consider the plane perpendicular to that line passing through the origin. consider orthonormal basis of that plane it will be a rotation in the plane due it the map being an isometry on the restriction and being orientation preserving while fixing the origin. Thus it is rotation about an axis passing through the origin.

I: Good.
I: What do you know about cardinality? Okay try this question for five minutes. I don`t think you will get it.
Consider V1 and V2 two vector spaces, let f: V1–>V2 be an onto linear map and
g: V2–>V1 be another onto linear map. Does there exist a linear isomorphism from V1 to V2?
A: (Proves for finite dimensional case that dimensions are equal which immediately leads to the result)
I: What about the infinite dimensional situation?
A: (Thinks for a couple of minutes.) I cannot say anything, it could be possible that both the vector spaces are uncountable of same cardinality while one has a countable basis.
I: Do you think two vector spaces with same cardinal numbers have bases of different cardinal numbers?
A: (?..)
I: Okay next question. List the sylow 2 subgroups of S3.
A: (Does so).
I: What can you say about groups of order 9?
A: Its a group of order p*p which is abelian, Hence by structure theorem the only possible groups of order 9 upto isomorphism are Z3xZ3 and Z9.
I: (Nods) Okay. So what is the madhava competition. You have got prize in it? How many people give the exam? Who organises it?
A: (Answers those questions)
I: Okay you may go now.
(Pick my bag and leave.)

After the interview I felt like I have done very little in the last three years.. I do not expect to get into TIFR Mumbai on the basis of this performance, fingers crossed though. I felt good that I could answer questions from topics I have studied.

UPDATE: Well the TIFR Ph.D. Selection for Mathematics is out. And surprisingly I have got in! I am pleasantly surprised by the outcome 😀 It will be a hard choice for me between Ecole Polytechnique and TIFR.