# Zero vector in a vector bundle

I am trying to get familiar with the idea of a vector bundle at an intuitive level, and it doesn't seem to be too abstract an idea: vector spaces get "attached" to points on a surface.

However, there is a naive, and early question that comes to mind reading the definition on Wikipedia:

... to every point x of the space X we associate (or "attach") a vector space V(x) in such a way that these vector spaces fit together to form another space of the same kind as X

It is one of the axioms of a vector space that there is an identity element of addition:

There exists an element $0 ∈ V,$ called the zero vector, such that $v + 0 = v$ for all v ∈ V

which I interpret to mean that vector spaces are centered at the origin of the coordinate system.

So in the case of a vector space attached to a point on some surface, is the zero the actual point?

• That's one way to think about it. It's particularly appropriate for thinking about the tangent bundle. Jul 29, 2017 at 0:11

For embedded manifolds certainly it makes a lot of sense to think of certain vector bundles as centring $V(x)$ at $x$. Examples:

The Cylinder: Consider $$X = \{(x,y,0) \in \mathbf{R}^3 : x^2 + y^2 = 1\}$$ (a circle) and $$E = \{(x,y,z) \in \mathbf{R}^3 : x^2 + y^2 = 1\}$$ (an infinite cylinder). Then $\pi : (x,y,z) \mapsto (x,y,0)$ makes $E$ into a 1-dimensional bundle (a line bundle) over $X$. The point $(x,y,0) \in X$ is also the corresponding $0$-vector of $\pi^{-1}(x)$.

The Mobius Strip (Image from Wikipedia) You can see the same picture with the point $x \in S^1$ corresponding to the $0$-vector of $\pi^{-1}(x)$. Maybe there is a small difficulty in viewing the entire infinitely wide Mobius strip at the same time, but we have a similar picture as with the cylinder.

The Tangent Bundles: For embedded manifolds, we can picture this in the same way as we did with the cylinder/Mobius strip. For non-embedded manifolds picturing it this way is a bit harder. For instance, consider the Klein bottle as the quotient of a square by identifying opposite sides with a twist: We can embed this in $\mathbf{R}^4$ but just from the square picture it is slightly more complicated to view the tangent bundle as being centred at each point as we did before. However if we just take a small open set in the Klein bottle, we can view that open set as a copy of $\mathbf{R}^2$ and draw our vector bundles, locally, as we would for $\mathbf{R}^2$.

In general, I think that the more abstract the vector bundle, the harder it is to picture it as being centred at each point. But to gain intuition, it is often helpful to draw pictures of the vector bundle in this way.

• Can you please introduce $\pi(\cdot)$? Jul 29, 2017 at 17:01
• @Antoni $\pi : E \to X$ is a vector bundle. If the elements of $E$ look like $(p, v)$ where $p \in X$ is a point of $X$ and $v$ is a vector in what you're calling $V(p)$ then $\pi(p, v) = p$. This is the notation used in the definition at Wikipedia. Jul 29, 2017 at 17:12
• I've edited a $\pi$ into the formula $\pi : (x,y,z) \mapsto (x,y,0)$ on the 7th line. Jul 29, 2017 at 17:14