Let $\imath :T^{n}\rightarrow X$ is a special Lagrangian n-Torus so that $\imath(T^{n})=L$ and all small special Lagrangian deformations of $L$ are flat then why $L$ has Tubular neighbourhood which is fibred by special Lagrangian Tori?
$\begingroup$
$\endgroup$
6
-
2$\begingroup$ @Hassan: By the way, I notice that you ask that "all small Lagrangian deformations" be flat. Does that mean 'flat in the metric induced by the immersion into the Calabi-Yau manifold $X$'? I assume you know that this cannot ever happen for $n>1$. Actually, when I wrote my answer below, I had ignored this assumption and assumed only that you meant for $L$ itself to be flat in the induced metric (which is all one needs for Maclean's theorem to give the local foliation result). I'm guessing that you are reading this somewhere and I suspect that the author of that article has been careless. $\endgroup$– Robert BryantCommented Feb 22, 2013 at 14:29
-
$\begingroup$ Dear Robert Bryant, Yes, In fact the metric induced by the immersion into the Calabi-Yau manifold $X$. I am not sure that we need to use of Tubular neighbourhood Theorem or not? I am studying Patric Baier thesis student of Hitchin, theorem 2.2.3 $\endgroup$– user21574Commented Feb 22, 2013 at 14:42
-
1$\begingroup$ @Hassan: But the point I'm trying to make is that you are describing an impossible situation. When $n>1$ it cannot ever happen that all of the small Lagrangian deformations of $L$ are flat in the induced metric. That is why I suspect that either you are misreading something or the author has been careless. By the way, I just downloaded Baier's 2001 thesis (people.maths.ox.ac.uk/hitchin/hitchinstudents/baier.ps.gz) and there is no Theorem 2.2.3 in it, so I can't see what you are talking about. $\endgroup$– Robert BryantCommented Feb 22, 2013 at 15:03
-
$\begingroup$ Thanks for your comment, I mean theorem 2.3.3, I had misprint in my previous comment, sorry $\endgroup$– user21574Commented Feb 22, 2013 at 15:32
-
1$\begingroup$ @Hassan: OK, so I can see that you left out a crucial adjective: He is not assuming that all small Lagrangian deformations are flat, only the small special Lagrangian deformations. (I can see that you have just added the word 'special' in your question, so you have now realized your error.) Now his proof goes through, and I don't see why you are asking the question. Are you not able to follow the proof? Note that in the discussion that follows, he points out that this 'semi-flat' situation will almost never hold. (I'm not convinced it can happen at all outside the flat case.) $\endgroup$– Robert BryantCommented Feb 22, 2013 at 15:40
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
1
NB: As the OP pointed out, the first version of my answer was incomplete, since it didn't address the 'tubular neighborhood' part of the claim. Here's a better (but still not complete) version of an answer
This is a partly consequence of Maclean's theorem. Since the $n$-torus itself is flat, a basis of the harmonic $1$-forms on $T^n$ are linearly independent at every point, so it follows from the description that Maclean gives in his theorem that the small deformations of $L$ as a special Lagrangian torus must be an $n$-parameter family that foliates a neighborhood $\mathcal{N}$ of $L$ in $X$. Let $\pi:\mathcal{N}\to B\subset\mathbb{R}^n$ be a submersion onto an open set $B$ in $\mathbb{R}^n$ whose fibers are these special Lagrangian leaves.
What's not obvious (and wouldn't be true without the hypothesis that the fibers are flat in the induced metric) is that $\pi$ is a Riemannian submersion with respect to an appropriate metric on the base $B$. The argument in Patric Baier's thesis (from which the OP drew the question in the first place) that the flatness of the fibers implies this is not completely obvious, but it follows from the normalizations that he has made on the previous page and the crucial observation that the differentials of the $x$-coordinates that he constructs on the fibers are harmonic with respect to the induced flat metrics on the fibers. This is what gives the constancy along the fibers of the coefficients $g^{ij}$ of the transverse metric. I don't see the point of reproducing this argument here; one should just go to Baier's thesis and follow that. It's available at http://people.maths.ox.ac.uk/hitchin/hitchinstudents/baier.ps.gz and the relevant arguments are on pages 48 and 49.
Once you know that the submersion is Riemannian, it follows immediately that any two leaves of the foliation are at constant distance from each other, so that the tubular neighborhood of one of the leaves is a union of leaves, which is what the OP wanted to know.
answered Feb 21, 2013 at 16:55
-
$\begingroup$ You explained first part and never explained why L has Tubular neighbourhood which is fibred by special Lagrangian Tori? $\endgroup$– user21574Commented Feb 22, 2013 at 14:34