# Vertical and horizontal subspace of a vector space T_pP.

1. Aug 24, 2010

### wdlang

suppose we have a principle fiber bundle P

at a point p \in P

we have the decomposition T_pP=V_pP + H_pP

it is said that the vertical subspace V_pP is uniquely defined while H_pP is not

i cannot understand this point

the complement to a unique subspace is surely unique, i think.

it is a basic fact in linear algebra.

2. Aug 24, 2010

### Bacle

I assume you mean a vector bundle, right.? (or maybe you use a different name for it),
and I guess you want a direct sum decomposition and uniqueness up to isomorphism.?

Where did you read that the complement was not unique (and unique up to what)?.

Then, you are correct: given an n-dimensional subspace S of an m-dim
vector space V , we can always define a subspace S' (e.g., by extending the
basis B_S of S to a basis B_V for V , so that S' is the subspace with basis
B_V-B_S )so that

V=S(+)S'

And dimensions add up, so DimS'=m-n . so S' is unique up to vector space isomorphism.

And all

3. Aug 24, 2010

### wdlang

no, it is about principle fiber bundle

it is in the context of horizontal lift of a curve in the base manifold M to a curve in the bundle P

it is said that some connect is needed to determine the horizontal subspace.

4. Aug 24, 2010

The orthogonal complement is unique, but that needs an inner product. In general, given a vector space V with a subspace U, there are many choices of subspaces W such that V is the direct sum of U and W.

5. Aug 24, 2010

### Bacle

But there is only one such W up to isomorphism, by invariance of dimension.

I have always been curious about "solving" for quotients, or direct sums, i.e.,

If we know V=S(+)S' , and we know S, how do we find S' up to isomorphism;

similarly, if we know that a group G is the quotient of two groups H,K, i.e.,

G~ H/K , and we only know either H or K, but not both, can we find the other.?

6. Aug 24, 2010

But the isomorphism problem for vector spaces is nearly trivial (two vector spaces are isomorphic iff they have the same dimension), making it uninteresting. If V = S ⊕ S', then S' is always isomorphic to V/S, but who cares?

Back to the original problem: It wasn't saying unique up to isomorphism; just unique.

edit: I did some reading about the objects in the original problem. A fibre bundle p: E -> M gives a canonically defined (i.e. uniquely satisfies a certain property) linear map of vector bundles Tp: TE -> TM; the vertical bundle VE is defined as the kernel of Tp. A horizontal bundle is then an arbitrary subbundle HE of TE such that TE = VE ⊕ HE. Even though any two horizontal bundles may be isomorphic, they may still be different subbundles of TE.

Generally speaking, if you're given a fixed object, you aren't interested in the various subobjects only up to isomorphism.

Last edited: Aug 24, 2010
7. Aug 25, 2010

### wdlang

to the point

thanks a lot!

8. Aug 25, 2010

### Fredrik

Staff Emeritus
Spelling nitpick: It's principal, not principle.

9. Oct 8, 2010

### math6

Re: vertical and horizontal subspace

where in the definition of vertical subspace we understand that the notion of canonical vertical vector: a vertical vector is a vector tangent to the fiber. ?

10. Oct 8, 2010

### lavinia

The vertical space is tangent to the fiber. But a complementary subspace of the tangent space to the fiber is not uniquely determined.