# Is there an Error on pg. 17 of Tromba's “Teichmuller Theory in Riemannian Geometry”?

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?)

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I realized that the answer to question 0) is "yes," or at least I see that the map $H \mapsto tr(J^{-1}H)$ is surjective. – BrainDead Jan 17 '11 at 4:43
Maybe I'm dense but isn't 1) just the Cayley-Hamilton theorem: $J^{2} - \text{tr}(J) \cdot J + \text{det}(J)\cdot\text{id} = 0$, so $J^{2} = - \text{id}$? – Theo Buehler Jan 17 '11 at 4:48
Why is $tr(J) =0$? – BrainDead Jan 17 '11 at 4:53
Sorry I confused $\mathcal{A}^{s}$ and $\mathcal{M}$. But then Cayley-Hamilton tells you that $J^{-1} = (\text{tr}(J)\text{id} - J)$, so $\text{tr}(J^{-1}H) = \text{tr}(J)\text{tr}(H) - \text{tr}(JH)$. – Theo Buehler Jan 17 '11 at 5:07

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 $T_J\mathcal{N}$.
Thank you for the answer. Yes, that should've been $\det^{-1}(1)$. I made the correction just now. But there is definitely a typo on pg. 17, since he says: "[after using $J^2=-id$ to get $Ddet(J)H = -tr(JH)$]...It remains to show that $H\mapsto -\tr JH$ is surjective. Then the implicit function theorem shows that $\mc{M}$ is a submanifold with tangent sapce $\ker(D\det(J))=\\{H|\tr JH =0 \\}.$ Thanks for the clarification. I thought perhaps I was missing some kind of a known theorem or some feature of the Implicit Function Theorem that allows me to only look at those with $J^2=-id$. – BrainDead Jan 17 '11 at 16:10
What about my question 2)? $\mathcal{A}^s$ is a $C^\infty$ submanifold of $\mathcal{N}$, and thus is a $C^\infty$ submanifold of $\mathcal{H}^s$? – BrainDead Jan 17 '11 at 16:12
In particular: the finite dimensional analogue gives if $M,N$ two smooth submanifolds of $X$ such that along $x\in M\cap N$ you have $T_xM$ and $T_xN$ span $T_xX$, then $M\cap N$ is a smooth submanifold of $X$ with codimension $\mathop{codim}(M) + \mathop{codim}(N)$. – Willie Wong Jan 17 '11 at 17:20