Representations of $S O( n )$ coming from $GL( N ,\mathbb{ R})$ I would like to show that "the finite-dimensional spinor representation of $SO(N)$ does not arise from a finite-dimensional representation of $GL(N)$" , as stated here and here.
Apparently we should take this to mean that you can't find/embed the double cover of $SO(n)$ in $GL(n,\mathbb{R})$.
(If needed, the original statement is here).
I am hoping what's done below is correct, any comments/fixes/illustrations/generalizations (even just in the comments section) would be greatly appreciated. I am not clear on the very end (posted a pic of that bit of the proof) and the generalization so comments on that would be helpful!
Consider first the group of rotations $SO(2)$, 
$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos(\theta) && - \sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$
of in the $(x,y)$ plane. We can explicitly derive the double cover by defining 
$$\begin{aligned} 
\zeta_0 &= \pm \sqrt{\frac{x-iy}{2}} \\ 
\zeta_1 &= \pm \sqrt{ \frac{-x-iy}{2} } 
\end{aligned},$$
they transform under rotations as 
$$\begin{aligned} 
\zeta_0' &= \pm \sqrt{\frac{x'-iy'}{2}} = \pm \sqrt{\frac{x \cos(\theta) - y \sin(\theta) - i x \sin(\theta) - i y \cos(\theta)}{2}} \\ 
&= \pm \sqrt{\frac{(x - i y)\cos(\theta) - i(x - iy) \sin(\theta)}{2}} = \pm  \sqrt{\frac{(x - i y)e^{-i \theta}}{2}} = e^{-i \theta/2} \zeta_0 \\ 
\zeta_1' &= \pm \sqrt{ \frac{-x'-iy'}{2} } = \pm \sqrt{ \frac{-x \cos(\theta) + y \sin(\theta) - i x \sin(\theta) - i y \cos(\theta)}{2} }  \\ 
&= \pm \sqrt{ \frac{-(x + iy) \cos(\theta) - i(x + i y)\sin(\theta) }{2} } = \pm \sqrt{ \frac{-(x + iy) e^{i \theta} }{2} } = e^{i \theta/2}\zeta_1
\end{aligned},$$
or
$$ \begin{bmatrix} \zeta_0' \\ \zeta_1' \end{bmatrix} = \begin{bmatrix} e^{-i\theta/2} && 0 \\ 0 && e^{i \theta/2} \end{bmatrix} \begin{bmatrix} \zeta_0 \\ \zeta_1 \end{bmatrix}$$ 
so that after a rotation by $2 \pi$ the signs change
$$ \begin{aligned} 
\zeta_0' &= \pm \sqrt{\frac{x'-iy'}{2}} = \pm \sqrt{\frac{e^{-i 2 \pi}(x-iy)}{2}} = \mp \sqrt{\frac{x-iy}{2}} \\ 
\zeta_1' &= \pm \sqrt{ \frac{-x'-iy'}{2} } = \mp \sqrt{ \frac{-x-iy}{2} }  \\ 
\begin{bmatrix} e^{-i\theta/2 - i \pi} && 0 \\ 0 && e^{i \theta/2 + i \pi} \end{bmatrix} &= \begin{bmatrix} - e^{-i\theta/2} && 0 \\ 0 && - e^{i \theta/2} \end{bmatrix} \end{aligned}.$$
In other words, to every $(x,y)$ there corresponds both a $(\zeta_0,\zeta_1)$ and a $(-\zeta_0,-\zeta_1)$, or  to every rotation matrix
$$\begin{bmatrix} \cos(\theta) && - \sin(\theta) \\ \sin(\theta) && \cos(\theta) \end{bmatrix}  $$
there correspond the two matrices
$$\begin{bmatrix} e^{-i\theta/2} && 0 \\ 0 && e^{i \theta/2} \end{bmatrix}, \begin{bmatrix} - e^{-i\theta/2} && 0 \\ 0 && - e^{i \theta/2} \end{bmatrix}.$$
Thus we have a 2-valued representation of the group of rotations, in the sense that one rotation $R$ in $SO(2)$ maps to two elements above. 
Now consider the elements of $A = \begin{bmatrix} a && b \\ c && d \end{bmatrix} \in SL(2,\mathbb{R}), \det(A) = 1$. We wish to know whether a representation of $A$ acting on a finite number of variables has the above 2-valued representation of $SO(2)$ as a sub-representation, that is, whether $SL(2,\mathbb{R})$ has multi-valued representations in the sense that $A$ in $SL(2,\mathbb{R})$ corresponds to multiple elements of a representation of $SL(2,\mathbb{R})$. 
If we were seeking to find multi-valued representations $\rho$ of $SL(2,\mathbb{C})$ then we would immediately see they do not exist since $SL(2,\mathbb{C})$ is simply connected. If we could find multi-valued representations of $SL(2,\mathbb{C})$ then defining a closed path $A(t)$ in $SL(2,\mathbb{C})$ we'd see $A(0) = A(1)$ while $\rho(A(0)) \neq \rho(A(1))$. On deforming the closed path down to a point, we have both $\rho(A(0)) = \rho(A(1))$ and $\rho(A(0)) \neq \rho(A(1))$ which is a contradiction. Since $SO(2)$ is not simply connected, we can't use this proof.
Analyzing $A \in SL(2,\mathbb{R})$ we see $\det(A) = 1$ allowing us to view elements as rotations and so we see the closed loops in $SL(2,\mathbb{R})$ are rotations about a circle thus the fundamental group (set of equivalence classes of closed loops) is $\mathbb{Z}$.
If we show the universal cover of $SL(2,\mathbb{R})$ with center $\mathbb{Z}$ is connected (in the page below) then this or it's representations cannot be used to obtain the above multi-valued representation. This shows that "the finite-dimensional spinor representation of $SO(2)$ does not arise from a finite-dimensional representation of $GL(2)$." 
For $N > 2$, I'm not sure why the above doesn't prove it by only focusing on two variables, indeed one source I read does this, but others seem to generalize it by modifying the end (e.g. the center is not $\mathbb{Z}$). 
Assuming the above is okay, this part is a bit unclear, the page it's taken from is below in case:
It seems that the way to do this is to analyze
$$y' = \frac{ay + b}{cy + d}$$
with $\det(A) > 0$ for $y \in \mathbb{R}$, set $y' = \tan(x') , y = \tan(x)$ and then differentiate 
$$\tan(x') = \frac{ a \tan(x) + b}{c \tan(x) + d}$$
to find regions of increase/decrease, then study the $x = 0$ case and shift the parameters by $\lambda$
$$\tan(x') = \frac{b + \lambda b_0}{d + \lambda d_0}$$
from $0$ to $\infty$ then $-\infty$ to $0$ to find the same $(a,b,c,d)$ give $x + \pi$ and repeating this shows to every $x$ we get an infinite number of $x'$'s.

 A: OK, here is one proof why $Spin(n)$ does not embed in $GL(n,R)$. Suppose that $\phi: Spin(n)\to GL(n,R)$ is any continuous representation. Since $Spin(n)$ is compact, so is its image. By Cartan's theorem, each compact subgroup of $GL(n,R)$ is conjugate into $O(n)$, see e.g. Mariano's answer here. 
Hence, by composing $\phi$ with conjugation, we can assume that $\phi(Spin(n))$ is contained in $O(n)$. Since $Spin(n)$ is connected, $\phi(Spin(n))$ is contained in $SO(n)$ (which is the identity component of $O(n)$). Since $Spin(n)$ is a simple Lie group and $\phi$ is 1-1, it follows that $\phi$ induces an embedding of Lie algebras $spin(n)\to so(n)$. But these Lie algebras are already isomorphic. Therefore, $\phi$ induces an isomorphism of Lie algebras. Therefore, $\phi: Spin(n)\to SO(n)$ is a covering map, see e.g.here. But then $\phi$ cannot be 1-1 since $Spin(n)$ is simply-connected and $SO(n)$ is not. qed  
Edit. Another way to argue is to compute dimensions of irreducible nontrivial spin-representations of the Lie algebra $so(n)$ and see that they are all larger than $n$. These representations are parameterized by dominant weights which are in the root lattice. However, this proof is less conceptual. 
