This is a very simple question that unfortunately requires a bit of an introduction

We have defined Manifolds in class from an analytical perspective as follows:

Take $1 \le q < n, k \ge 1, V \in R^n, V \ne \emptyset$.

Then we call $V$ a $q$-manifold of class $C^k$ if $\forall x_0 \in V \ \exists \ B(x_o,\delta), \ f: B(x_0, \delta) \rightarrow R^{n-q}$ s.t.

  1. $\{ x \in B(x_0, \delta) | f(x) = 0 \} = B(x_0, \delta) \cap V$

  2. $f \in C^k$ (the set of all continuous functions with continuous partial derivatives up until the $k$th)

  3. rank of the Jacobian of $f$ is $n-q$ $\forall x \in B(x_0, \delta)$

We also stated a theorem that says that if $V \subset R^n$ is a q-manifold of class $C^k$ and if $x_0 \in V$ then the set of vectors $h$ such that $h$ is tangent to $V$ in $x_0$ is the $Ker(df(x_0)$ where $df(x_0)$ is the matrix associated with the total differential.

I wanted to try out this theorem with a very simple example, as $V$ I take line that bisects the first quadrant, the $f$ of the definition of $V$ will be $f(x,y) = x-y$ correct?

But now the total differential is represented by the vector $(1,-1)$ and the Kernel is $$<(1,-1), (h_1, h_2) >$$ \implies $h_1 = h_2$ But the vectors of the form $h_1 = h_2$ are parallel not tangent to the line, where is my mistake?

  • $\begingroup$ I don't understand your confusion. The vector $(h_1, h_1)$ is tangent to $V$. $\endgroup$ – Michael Albanese Jan 13 '16 at 11:01
  • $\begingroup$ @MichaelAlbanese Thank you for your answer Michael, would I not want a vector of the type $( - h_1, h_1)$? Because $V$ is the set of points that bisect the first and third quadrant? $\endgroup$ – Monolite Jan 13 '16 at 11:25
  • 1
    $\begingroup$ A vector of that type will be normal to $V$, not tangent to it. $\endgroup$ – Michael Albanese Jan 13 '16 at 11:26
  • $\begingroup$ @MichaelAlbanese em thanks a lot I get it now obviously you are right, I am happy I cleared this. $\endgroup$ – Monolite Jan 13 '16 at 11:33

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