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 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?)