# counstruction of vector bundle on projective space

I have a question about real projective space $\mathbb{R}P^n$. I am studying such an example on construction a vector bundle, but I am not familiar with the projective space at all.

We have $X=\mathbb{R}P^n$ and let $L\subset X\times \mathbb{R}^{n+1}$ denoted by $$L=\left\{(l,v) : l\in X, v\in l\right\}.$$

$L$ is a subfamily of the trivial family and our goal is to show that it is a line bundle over $X$. So we want to produce a local trivialization for any $l\in X$. WLOG we concider the case when $l=<e_1>$. Now it starts be harder for me. Let $U\subset\mathbb{R}P^n$ be the set of lines whose orthogonal projection to $<e_1>$ is nonzero. Such a line contains a unique vector of the form $e_1+u$, such that $e_1\cdot u=0$ (why?). Then define $s:U\rightarrow L$ by sending $l$ to $(l,e_1+u)$ where $e_1+u$ is the unique point on $l$ described above. This section is clearly nonzero everywhere (why?), so it gives a trivialization of $L|_{U}$. Thus, we have proven that $L$ is locally trivial and hence a vector bundle.

If someone could comment the whole construction, specially places which I marked as confusing, I would be greatful.

• You say $L\subset X\times \mathbb{R}^{n+1}$ and then $L=\left\{(l,v) : l\in X, v\in\mathbb{R}^{n+1}\right\}.$ If you allow $v$ to range over all vectors in $\mathbb{R}^{n+1}$, then you actually have $L$ is all of $X\times \mathbb{R}^{n+1}$, rather than a proper subset. It's not a line bundle at all, it's an $(n+1)$-bundle. Surely what you meant to say is $L=\left\{(l,v) : l\in X, v\in l\right\}$? This is sometimes called the tautological line bundle. Nov 20, 2017 at 12:52
• Sure, you are right. Nov 20, 2017 at 13:14

As mentioned by ziggurism in the comments you probably meant to write that $v\in l$ instead of $v\in \mathbb{R}^{n+1}$. Otherwise the bundle is simply the trivial bundle of rank $n+1$ on $X$ and thus also locally trivial. So let's assume you meant

$$L=\{(l,v)\mid l\in X, v\in l\}.$$

So in words $L$ consists of the pairs of a line in $\mathbb{R}^{n+1}$ and a point on that line. The bundle projection is simply $(l,v)\mapsto l$ and the fiberwise additions and scalar multiplication are $\lambda(l,v)=(l,\lambda v)$ and $(l,v)+(l,w)=(l,v+w)$ where we not that the sum of two vectors on a line through the origin is still on that line through the origin thus the addition is well-defined.

Now for the part that you seem to struggle with, the local trivializations. First note that the fiber of the $L$ over $l\in X$ is given by $l\subset \mathbb{R}^{n+1}$ so that the dimension of the fibers is $1$. Thus a trivialization is given by a non-vanishing section. By changing the basis of $\mathbb{R}^{n+1}$ we obatin a trivialization around any line $l\in X$ if we obtain one around the line $\langle e_1\rangle\in X$ spanned by $e_1$.

Now consider the subset $U\subset X$ of lines $l\in X$ which are not orthogonal to $\langle e_1\rangle$. Another way to characterize $U$ is that $l\in U$ if and only if $v\cdot e_1\neq 0$ for all $0\neq v\in l$. Suppose $l\in U$ and pick some $0\neq v_0\in l$. Note that for all $v\in l$ there is $\lambda_0\in \mathbb{R}$ such that $v=\lambda v_0$ (since $l$ is a line). Set $$w=\frac{v_0}{v_0\cdot e_1}$$ and note that $w$ does not depends on $v_0$, since if we had started with $v_1=\lambda_1v_0$ then $\frac{v_1}{v_1\cdot e_1}=\frac{\lambda_1v_0}{\lambda_1v_0\cdot e_1}=\frac{v_0}{v_0\cdot e_1}$. Note also that $w\cdot e_1=\frac{v_0\cdot e_1}{v_0\cdot e_1}=1$. Now if we write $w$ in the orthogonal basis $(e_1,\ldots, e_{n+1})$ as $$w=\sum_{i=1}^{n+1}a_ie_i$$ we see that $1=w\cdot e_1=a_1$ and so indeed $w=e_1+u$ with $e_1\cdot u=0$ (where $u=\sum_{i=2}^{n+1}a_ie_i$). It is clearly the unique vector of the form $e_1+u'\in l$ with $e_1\cdot u'=0$ since if we set $e_1+u'=v_0$ we find $$w=\frac{v_0}{e_1\cdot v_0}=e_1+u'.$$

Now set $s\colon U\rightarrow L$ to be the map $s(l)=(l,w)$ then this map is clearly a section and $w=0$ implies $1=e_1\cdot w=0$ which is a contradiction. So we see that $s$ is a non-vanishing section and thus trivializes $L$ over $U$.

I hope this clarifies the construction a bit for you, if questions remain please put them in the comments or edit your question above.

• I get it now, thank u Nov 20, 2017 at 19:57

So $U$ is the set of lines not perpendicular to $e_1.$ Your first question is, why does any line not perpendicular to $e_1$ contain a vector of the form $e_1+u$ with $u\cdot e_1=0?$

Let's recall what the vector dot product does; it tells you the components of orthogonal decomposition. For unit vector $e_1$ and any vector $v$, we may decompose $v$ into parts parallel to $e_1$ and perpendicular, by writing $v=v_\parallel+v_\perp=(v\cdot e_1)e_1+(v-v_\parallel)=(v\cdot e_1)e_1+\left(v-(v\cdot e_1)e_1\right).$

The line $\mu$ spanned by $v$ is the set of all $\mu = \{\lambda v\colon\lambda\in\mathbb{R}\}.$ (I am calling this line $\mu$ because we are already using the letter $\ell$ for $\langle e_1\rangle$, and I don't want to get confused.)

If we assume that $\mu$ is not perpendicular to $e_1$, then $v\cdot e_1\neq 0.$ So in particular one vector in the span of $v$ is $$\frac{v}{v\cdot e_1}=\frac{v_\parallel+v_\perp}{v\cdot e_1}=\frac{(v\cdot e_1)e_1+v_\perp}{v\cdot e_1}=e_1+\frac{\left(v-(v\cdot e_1)e_1\right)}{v\cdot e_1}=e_1+u.$$

To sum up, orthogonal decompositions are unique, and if the vector we started with had a nonzero $e_1$ component, then it can always be scaled to a unique vector with $e_1$ component exactly 1. So every line not perpendicular to $e_1$ has a unique vector of the form $e_1+u$ with $e_1\cdot u = 0.$

Next question is: why is the section $\mu\mapsto e_1+u$ nonzero? Well to get the zero vector, we'd have to have $e_1=-u$. But since these vectors are components of an orthogonal decomposition, they are never colinear. They never sum to zero.

Can I say anything to make this construction more understandable?

Think of the $n=1$ case. $\mathbb{R}P^1$ is a circle. Lines in the plane through the origin. So the unit circle in the plane, but with diametrically opposed points identified, since the same line goes through them. At each point on the circle, there is the line through that point. This makes a trivial bundle, but by identifying opposite points, you introduce a twist. It's the Möbius band, viewed as a real line bundle. And $(\cos\theta,\sin\theta)$ is not perpendicular to $e_1$ if $\theta\neq \pi/2.$ So our nonvanishing section of the bundle over $\{0\leq\theta<\pi/2\}$ is then $\theta\mapsto (1,\tan\theta).$

• thank you, it was helpful Nov 20, 2017 at 19:59