I'm pretty sure that this is a minor error, but I could use some help here. So the book I'm referring to in the title is [this](https://www.springer.com/us/book/9783764327354) book (MR1164870).


On pg. 16-17, he is proving that the space of almost complex structures on a compact smooth surface without boundary is a smooth submanifold of the (1,1)-tensors. He does by expressing the space of almost complex structures is the intersection of traceless tensors and the tensors of determinant 1. He then uses transversality to conclude the argument.


I don't understand Tromba's proof of the statement that the space of tensors of determinant 1 is a smooth submanifold of the (1,1)-tensors. He says he uses the Implicit Function Theorem, but what he does doesn't really make sense to me.


I will use the notations that he uses in the book and present his argument, or at least my poor understanding of his argument.


Let $\mathcal{H}^s(T^1\_1 M)$ denote the space of (1,1) tensors on a smooth surface $M$ of some genus $g$, which are weakly differentiable to order $s$ and are square integrable. (Let $s>3$.) 


Let $\mathcal{A}^s = \{J\in \mathcal{H}^s| J^2=-id, \text{for any } x\in $M$, v\in T\_x M, (v,Jv)\text{form an ordered basis} \}$. This is the space of almost complex structures.


Let $\mathcal{N}:=tr^{-1}(0)\subset\mathcal{H}^s(T^1\_1 M)$. This is a linear subspace, thus is automatically a smooth submanifold.


Let $\mathcal{M}:=det^{-1}(1) \subset \mathcal{H}^s(T^1\_1 M)$. It's easy to check that $\mathcal{A}^s = \mathcal{N} \cap \mathcal{M}$.


Claim: $\mathcal{M}$ is a $C^\infty$ submanifold of $\mathcal{H}^s(T^1\_1 M)$ with $T\_J\mathcal{M} =\{ H | tr JH = 0\}$


To use the Implicit Function Theorem, we want to show that $Ddet(J):T\_J \mathcal{H}^s \to \mathbb{R}$ is surjective for every $J \in \mathcal{M}$.


It's easy to see that $Ddet(J)H = (det J) tr(J^{-1}H) = tr(J^{-1}H)$.


But now, Tromba goes onto say that $J^{-1} = -J$, and concludes that $Ddet(J)H = -tr(JH)$. From this expression, the surjectivity is obvious. But this doesn't make any sense, since $J$ is just an element of $\mathcal{M}$, and there is nothing in the definition of $\mathcal{M}$ that forces $J^2=-id$!!!


My questions are, 


0) Is $\mathcal{M}$ a $C^\infty$ submanifold of $\mathcal{H}^s$, regardless of what is written here?


1) Why does he get to choose a $J$ such that $J^2 = -id$? Why is that sufficient to prove that $\mathcal{M}$ is a submanifold?


2) This argument does show that $\mathcal{A}^s$ is a $C^\infty$ submanifold of $\mathcal{N}$. Does that imply that $\mathcal{A}^s$ is a $C^\infty$ submanifold of $\mathcal{H}^s$? (i.e. is being a "$C^\infty$ submanifold" transitive?)


Thank you for your time. And I apologize if this is not an appropriate question here.


EDIT: I think some people misunderstood my question 2). What I was asking was whether transversality is necessary at all.


So we know that $\mathcal{A}^s=\det^{-1}(1)\subset\mathcal{N}$. Using IFT, we can conclude that $\mathcal{A}^s$ is a $C^\infty$ submanifold of $\mathcal{N}$, the same way one could conclude that $\mathcal{M}$ is a $C^\infty$ submanifold of $\mathcal{H}^s$. 


Does it follow from this that $\mathcal{A}^s$ is a smooth submanifold of $\mathcal{H}^s$? (That is, without saying anything about transversal intersections?)



Is there an Error on pg. 17 of Tromba's "Teichmuller Theory in Riemannian Geometry"?
A small typo: your definition of $\mathcal{M}$ is incorrect; shouldn't it be $\det^{-1}(1)$? 


Anyway, I think you are mixing up two parts of the proof. The first is that $\mathcal{M}$ is a smooth submanifold. Indeed, using your computation, it is easy to see that taking $H= J$ you have $D\det(J)H = \mathop{tr}(J^{-1}J) = 2$ and so surjectivity follows. 


The part about setting $J^{-1} = -J$ is only used to prove the transversality of $\mathcal{M}$ and $\mathcal{N}$ at the intersection: that if $J\in\mathcal{M}\cap\mathcal{N}$, then indeed $D\det(J)H = -\mathop{tr}(JH)$. In particular for such $J$ you have $Id\in T\_J\mathcal{M}$ but not in $T\_J\mathcal{N}$.


