I am reading Warner's Differentiable Manifolds I do not get one example which is

Let $V$ be a finite dimensional real vector space. Then $V$ has a natural manifold structure. If $\{e_i\}$ is a basis then the elements of the dual basis $\{r_i\}$ are the coordinate functions of a global coordinate system on $V$.

I don't understand how "the elements of the dual basis $\{r_i\}$ are the coordinate functions of a global coordinate system on $V$." Could any one explain me about that? Then how such a global coordinate system uniquely determines a differentiable structure on $V$? And why this structure is indipendent of choice of basis?

First of all for a manifold structure I need each point must have an open neighborhood $U$ homeomorphic to some open subset of $\mathbb{R}^n$. Here am I getting such notions?


1 Answer 1


The space $\mathbb{R}^n$ has coordinate functions $x_j:\mathbb{R}^n\to\mathbb{R}$, projection onto the $j^{th}$ axis. If $(\phi,U)$ is a coordinate system on a manifold $M$, then we get coordinate functions on $U$ by composing $\phi$ with the $x_j$.

Warner is just saying that by choosing a basis on a real vector space $V$, you induce a bjiective linear map (hence homeomorphism) $A$ between $V$ and $\mathbb{R}^n$, and that homeomorphism is a global coordinate system with coordinate functions $x_j\circ A = r_j$. The open neighborhood about each point is the entire space $V$.

To see that the structure is independent of choice of basis (up to diffeomorphism), try the construction with a different basis; can you see a diffeomorphism between the two structures?

  • $\begingroup$ Thank you Neal, But could you tell me how a nbd of a point is the whole vector space $V$? $\endgroup$
    – Myshkin
    Jun 19, 2012 at 13:04
  • 1
    $\begingroup$ A neighborhood of a point is an open set containing the point. The whole vector space is an open set, so it is a neighborhood of any point in it. $\endgroup$
    – Neal
    Jun 19, 2012 at 13:26
  • $\begingroup$ In which way is the equipment of manifold structure on a vector space independent on the choice of the basis? For a "pure" vector space (without inner product) the morphism $V\rightarrow V^*$ is not intrinsic. Could you clarify this point? $\endgroup$
    – cards
    Sep 27, 2022 at 15:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .