Third Year Revision Thread

Wow! But enough to lose alphas? If you throw enough questions at a wall some of them will get alphas I mean, I don't think I've done a question yet where I've not made a small error here or there but I'm confident of at least 5 alphas so far (note to self: get at least 8 more).



Indeed.

Did people moan about Part II exams last year as much as we are this year? I can't find the relevant thread.

I'm just hoping I do well enough to get into Part III, 'cos I really want to be here next year.

I've already put down a deposit to rent a house...

Snap. Results come out on my birthday. Planning on leveraging that when guilt tripping my DoS (if necessary)

If you get a 2:1 then the likes of me and around are doomed!

and again

Wow! But enough to lose alphas? If you throw enough questions at a wall some of them will get alphas

These exams are a such a bad joke. As someone already said, three years ending in farce.
As a suggestion for future reference: it would make sense to put section I questions for the D courses as well, as a little acknowledgment; it's really frustrating otherwise. Also, the disparity in difficulty across the courses is quite disconcerting. The whole system is a joke. Sorry for the rant, not a happy bunny these days.

Far from ideal, but a lot better today.

I missed the Topics in Analysis question that I could definitely do without having been to the lectures. Dammit!

I messed up Rep Theory because I thought it meant that it was talking about the representation on $\mathbb{C}X$ rather just $G=\text{Sym}(X)$. But I think Logic & Set Theory, Galois and Algebraic Topology went okay. (Out of interest, what did other people get for the homology groups of the 3-sphere-minus-a-solid-torus? I got $H_n(M) \cong \mathbb{Z}$ for $n=0,1$ and 0 for higher homology groups.)

Hmmm. Quick calculation here says that $H_n (M)$ should be the free abelian group of rank 1 for n = 0, 1. But above that it's not so clear. We have an exact sequence

so in principle $H_3 (M)$ could be a free abelian group of rank 0 or 1, and $H_2 (M)$ could be any finite cyclic group. It all depends on what the connecting map $H_3 (S^3) \to H_2 (S^1 \times S^1)$ does, and that requires more geometric intuition than I have available...

. (Out of interest, what did other people get for the homology groups of the 3-sphere-minus-a-solid-torus? I got $H_n(M) \cong \mathbb{Z}$ for $n=0,1$ and 0 for higher homology groups.)

I am pretty sure that this is correct since the 3-sphere minus a solid torus is just a solid torus and then you have $H_n(M) \cong H_n(S^1 \times D^2) \cong H_n(S^1)$.

To see that $M$ is a solid torus I would think of the Hopf Fibration $S^1 \hookrightarrow S^3 \to S^2$. Take a copy of $D^2$ on $S^2$. Pulling back, by local triviality, we get $D^2 \times S^1$. Since the complement of $D^2$ in $S^2$ is $D^2$; it follows that $D^2 \times S^1$ is complemented by $D^2 \times S^1$ in $S^3$.

Ah right; fortunately I don't think we were expected to know that a 3-sphere minus a solid torus is another solid torus. (It would make the question very easy if we knew that.)


If you say so. Maybe in a Part III exam!