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Let $A:V \to V$ a Linear Operator on the Vector Space $V$ and $Id:V \to V$ the Identity map, when is $A$ transverse to $Id$ ? My intuition says that $A$ may not have $1$ as eigenvalue but I didn't find out if it is really true

Since

Two maps $f: X \to Z$ and $g:Y\to Z$ are transversal if, for every $ x \in X$ and $y\in Y$ with $f(x) = z = g(y)$, the diferentials on these points spans the entire tangent space at $z$ in sense that $$im(df) + im(dg) \simeq T_zZ $$ (by https://ncatlab.org/nlab/show/transversal+maps)

We have

$A: V \to V$ and $Id:V\to V$ are transversal if, for every $x \in V$ and $y\in V$ with $A(x) = z = Id(y)$ , the diferentials at these points spans the entire tangent space at $z$ in sense that $$im(A) + im(Id) \simeq T_zV \simeq V$$ , i.e.

$$A(x) + Id(y) = z \ \ \forall \ \ x, y \in V \ s.t. \ \ A(x) = z = Id(y) , \ i.e. \\ \iff (A - I)(x) \ = x \ \ * \ for \ x \ = y \ = z$$

Since $(A-\lambda I)$ is singular $\iff det(A-\lambda I)=0$ $\iff \lambda $ is eigenvalue of $A$ , if $1$ is eigenvalue of $A$ then equation $\ \ * \ \ $ doesn't means isomorphism and then, by if and only if, doesn't means transversality of $A$ and $Id$.

My first problem is:

I assumed at equation $\ \ * \ \ $ the equality $x \ = y\ \ =z$, which makes sense to me because $A$ and $Id$ are global maps and then the transversality must be with all points of $V$ but I'm feeling that something is missing. $z \ = y \ $ because $Id$, no doubt, but $ x \ = y \ $?

My second problem is:

At https://ncatlab.org/nlab/show/transversal+maps the author says

'In particular, a submersion is transversal to all functions.'

But isn't Identity and non-singular Linear Operator surjective , equals to they derivative , because they are their best linear approximation, and then are submersions ?

BraQuiet
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  • The original question I'm trying to solve is that :

    https://math.stackexchange.com/questions/3147047/the-diagonals-delta-v-v-mid-v-in-v-is-transversal-to-w-v-av-mid-v-in

     – BraQuiet Dec 11 '19 at 14:22

1 Answers1

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Submersions are surjective on tangent spaces, by definition. This means the image of $df$ is the entire tangent space. In particular, no matter what other subspace you add to it, you will get the entire tangent space. This means it is transverse to everything.

I am not sure you are asking the question you intend to ask, however, since transverse intersection is a concept that applies to maps between manifolds, and yet you started out with just a linear operator on a vector space. Are you intentionally treating this vector space as both a manifold and identifying it with the tangent spaces? Or are you doing something else? If this is a homework question, I suggest you state this and post the question as close to verbatim as possible to avoid introducing notational mistakes.

pre-kidney
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  • I intended to write the least possible to not take so much time of fellow reader and expose the core of my problem with this verification problem.

    There is some notational mistake in what I wrote ? I'm learning and would like to know

     – BraQuiet Dec 11 '19 at 07:33
  • I pointed out two problems I've found in my solution – BraQuiet Dec 11 '19 at 07:37
  • Yes, your notational mistake is that the way you are using the word transverse doesn't seem logically like what the question is asking. My guess is that you are using the slightly wrong word (or mistranslation from a foreign language) and you mean something more like "orthogonal" instead of "transverse". Then you looked up the definition of transverse and it happened to take you to ncatlab where you found the definition of a totally different concept for a manifold. – pre-kidney Dec 11 '19 at 07:41
  • But all this would be clarified if you posted the context of your original question. I know what you are trying to do, to make it a simpler question, but I think your mistake is actually in that step of transferring the question to the form for this website. Is the original question in another language? – pre-kidney Dec 11 '19 at 07:42
  • One more reason it is a little more confusing is that the totally unrelated definition from ncatlab can be made to apply, but that is just a misleading coincidence. – pre-kidney Dec 11 '19 at 07:43
  • Good guess, could be the case, but I am studying the book of Differential Topology written by Guilemin and Polack and I mean 'transverse' as is in the book. In the ncatlab I found this clear definition and choose to put here – BraQuiet Dec 11 '19 at 07:47
  • This exercice I made as Lema to proof another resut they ask in the book and as verification problem to me to understand if I understood – BraQuiet Dec 11 '19 at 07:48
  • Okay, this exercise doesn't make much sense then. I guess it would make sense if you did something like: let $f\colon X\to X$ be a function with fixed point $p\in X$, then replace $V$ with $T_pX$ and replace $A$ with $df_p$ and replace $I$ with the differential of the identity map from $X$ to itself. – pre-kidney Dec 11 '19 at 07:55
  • Thx

    If we look to $V$ as manifold with global chat as $(V,Id)$ , $f=A$ , we wouldn't have $df = A $ and $dI =I$ ?

     – BraQuiet Dec 11 '19 at 08:03
  • The original problem, in short, is :

    Let $A:V\to V$ a linear map and $\Delta :V \to V\times V$ the diagonal application. Show that $ W = (V,AV)$ is transversal to $\Delta$ if, and only if, $1$ is not an eigenvalue of $A$

     – BraQuiet Dec 11 '19 at 08:06
  • In that case we have $$Im(dA(V)) = A(V), Im(dI(V)) = I(V)$$ and then $$Im(dA)+Im(dI) \simeq A(V) + I(V) \ \simeq T_pV \simeq V \ \sim \ A(V)+I(V) \ = V $$

    Edit: In the case we assuming the problem you suggested with f = A

     – BraQuiet Dec 11 '19 at 08:12