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Why every smooth orientable 4-dimensional manifold admits an immersion into $\mathbb{R}^{6}$?

This is a one-line question as I see the statement in a comment by Michael Hopkins(update: this is wrong, it is by Peter Kronheimer) . I thought about it for a long time but I do not know how to prove it or to approach it. Characteristic classes provide a way of showing "if this...", but does not help to show the existence of such an immersion. (comment: this is stupid line of thinking because obviously I did not make use of all characteristic tools available to me, as evident in reading Kirby's book)

update: The correct statement is provided in Kirby's book, page 44 Lemma 1, which states such immersion exists iff there exists a characteristic class $x\in H^{2}(M^{4};\mathbb{Z})$ such that $x_{(2)}=-w_{2}$ and $x^{2}=-p_{1}$.

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This is probably overkill but there's a result of Cohen saying that every $n$-manifold immerses into a $S^{2n-\alpha(n)}$ where $\alpha(n)$ is the number of $1$'s in the binary expansion of $n$. Now $\alpha(4) = 1$, so every $4$-manifold immerses into $S^7$ and since such an immersion won't be onto, you can project stereographically to $R^6$. Here's Cohen's paper. – t.b. Nov 14 '11 at 22:36
@t.b.- wouldn't that be $\mathbb{R}^7$? – user641 Nov 14 '11 at 23:02
@t.b.: I am aware that this might have generalization like every $2n$ manifold have an immersion into $R^{3n}$, etc. But it feels a mile away as I am only attacking on a small problem. I will try to read Cohen's paper in my sparetime. As the previous commentor noted, stereographical projection only offer one from $\mathbb{S}^{n}$ to $\mathbb{R}^{n}$. In your case it should be $\mathbb{R}^{7}$. – Kerry Nov 14 '11 at 23:35
@SteveD: sure, you're absolutely right, this was just plain silliness on my part, I leave the comment despite its stupidity. Thanks for the correction :) – t.b. Nov 14 '11 at 23:46
t.b. uh oh, now I have to think about this! – Ryan Budney Nov 15 '11 at 1:54
up vote 6 down vote accepted

This is the main topic of Chapter VI (entitled "Immersing $4$-manifolds in $\mathbb{R}^6$") of Kirby's book "The Topology of 4-Manifolds".

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This comment is very helpful. I found Kirby's book and his discussion is clear and understandable. Thanks a lot for the information, I really do not know much about 4-dimensional manifolds. – Kerry Nov 15 '11 at 4:15
Kirby's book is a fantastic source for this kind of stuff. I also recommend Scorpan's book "The Wild World of 4-Manifolds". – Adam Smith Nov 15 '11 at 4:21

I suspect you either mis-read the Hopkins statement, or Hopkins made an error. Likely the statement he intended is that every orientable 4-manifold is cobordant to one that immerses in $\mathbb R^6$ -- this is in Kirby's book on 4-manifolds and is a commonly used step in the proof of Rochlin's theorem.

But not every orientable 4-manifold immerses in $\mathbb R^6$, as $w_4$ is an obstruction. This is a theorem of K. Sakuma's. Sakuma Reference

So for example, $\mathbb CP^2$ does not immerse in $\mathbb R^6$ according to Sakuma, since it has odd Euler characteristic.

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I think your suspicion is justified. I did not mis-read Hopkins' statement; it is in here (…). Your explanation is very clear. So thank you a lot for clearing my thoughts. Kirby's book give an if and only if condition for such an immersion to exist, and obviously $w_{1}$ is not enough. I shall update my post. – Kerry Nov 15 '11 at 4:41
A great thing about being a prof at MIT is you can feel comfortable giving students homework problems where they're asked to prove false statements. It keeps those students on their toes, and (as a prof) you get to find out which ones are sharp! I don't have the Kirby book at home -- could you include Kirby's if and only if statement? – Ryan Budney Nov 15 '11 at 4:47
I updated the question contents. I am a student in Bard, not in MIT, so I feel sympathetic for Hopkins's real students. I know I am not brave enough to challenge a master by finding a counter-example myself. I need some time to digest Kirby's proofs. You may find his book in here: (…) – Kerry Nov 15 '11 at 4:58
@Adam Smith: it may actually be an error of someone else, as the files showed the signature of Peter Kronheimer, who is another professor in Harvard. So Hopkins may just make use of his HW sheets. I would not pursue this line of thinking as it is obviously unhelpful to help achieve a better understanding of this problem. I will try to write an essay on Kirby's proof and post it in here. – Kerry Nov 15 '11 at 5:34
Nice. I'm happy to see that you were up to the challenge I carelessly imposed on you :) – t.b. Nov 15 '11 at 6:04

I emailed Michael Hopkins and he said this problem set is not his, but composed by Peter Kronheimer. So the content of my question is quite inappropriate and I apologize for the misnomber in here. I will leave the question unchanged (for otherwise the answers may be incomprehensible). I will update this "answer" once I worked out Kirby's proof.

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