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I guess I am missing something obvious here. I am reading about vector bundles. (What Karoubi calls 'Quasi Vector-Bundles')

An example is the sphere, where for every point $X \in S^n$ we choose $E_X$ (the fiber) to be the vector space orthogonal to $X$, and let $E$ (the total space) be the disjoint union of the $E_X$'s, which is naturally a subspace of $S^n \times R^{n+1}$

Now later it states we can have an (iso)morphism to the space E'= S^1 \times \mathbb{R} given by $g(x,z) = (x,iz/x)$

Now I must be missing something here. If $x \in S^1$ and $z \in \mathbb{C} = \mathbb{R}^2$ (identifying $\mathbb{R}^2$ with the complex numbers), then doesn't $g(i,a+i)=(i,a+i)$, and $a+i \notin \mathbb{R}$.

What am I missing? $a+i$ is in the vector space orthogonal to $i$ (which is just the tanget line at the point $(0,1)$ of the circle)

(Feel free to re-tag, wasn't sure what it should be under)

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    Yes. The tangent space $T_x M$ at a point $x$ is by definition a vector space. What you're thinking of is called the *affine tangent space* $x + T_x M$ of an embedded submanifold $M$ of $\mathbb{R}^{n+1}$.2011-05-02

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The point is that $\{a + i\,:\,a \in \mathbb{R}\}$ (the affine tangent space of $S^1$ at $i$) is not a vector space. The tangent space $T_x S^1$ can be identified with $\{iax\,:\,a \in \mathbb{R}\}$, so for $z = iax \in x^{\perp}$ we have $iz/x = -a \in \mathbb{R}$.

More generally, the affine tangent space of an embedded submanifold $M$ of $\mathbb{R}^{n}$ at a point $x \in M$ is $E_x M = x + T_x M$. Geometrically, $E_x M$ is the affine subspace of $\mathbb{R}^n$ tangent to $M$ at the point $x$, but intrinsically only the vector space $T_x M$ makes sense (the space $E_x M$ depends on the embedding $M \hookrightarrow \mathbb{R}^n$).