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?



 special Lagrangian n-Torus has Tubular neighbourhood?
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.


