I have to show that the projective space $\mathbb{P}^n$ is isomorphic to the grassmanian $\mathrm{Grass}(1,n+1)=\{V\subseteq\mathbb{R}^{n+1}:V\,\text{linear subspace,}\,\dim\,V\,=1\}$ as well as describe the sets $\mathbb{P}^n-U_0$ and $\mathbb{P}^1-U_0$("difference of sets"), where $U_0$ is defined as $U_0:=\{[x_o:\,...\,:x_n]:x_0\neq0\}$ and state wheather these are manifolds or not. These questions might be very easy but I am a first year mathematics student so I do not have that much experience, as I have only taken analysis 1 and linear algebra 1 so far. Thank you very much in advance for your help!

  • $\begingroup$ Projective space is the space of lines, the Grassmannian manifold is the same space by definition. You can look at Pl\"{u}cker coordinates for both to see they define the same submanifold of some $\mathbb{R}^n$. It basically amounts to saying the words "they have the same definition, so they are the same." Unless you're using some other definition for projective space, maybe? $\endgroup$ – Adam Hughes Jul 9 '14 at 22:23
  • $\begingroup$ I use the following definition;$\mathbb{P}^n:=(\mathbb{R}^{n+1}-\{o\})$/~ where the relation is defined by $(y_0,...,y_n)$~$(x_0,...,x_n):<=>y_i =\alpha_i x_i$ for all i=0,...,n . $\endgroup$ – Hans Jul 9 '14 at 22:31

In light of your comment you can produce the isomorphism by explicitly exhibiting coordinates on $\mathbb{P}^n$ and noting that they are the same coordinates as those for Grassman (1, n) space.

Claim: Pl\"{u}cker coordinates are coordinates on projective space

Proof: a line through the origin in $\mathbb{R}^{n+1}$ is given by a vector equation $\ell: \{t\mathbf{v}:t\in\mathbb{R}\}$ for some non-zero vector $\mathbf{v}$ which. But then this is exactly the same equivalence relation put on $\mathbb{R}^{n+1}\setminus\{0\}$ in the question, hence


with $$\mathbf{v}\sim\mathbf{u}\iff \mathbf{v}=t\mathbf{u}$$

for some $t\in\mathbb{R}\setminus\{0\}$.

So the isomorphism is given by sending $x\in\mathbb{P}^{n}$ to any point on the line through the origin that it represents when you think of it as a vector in $\mathbb{R}^n$. The equivalence relation $\sim$ shows the map is well defined.


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