1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Path Components

  1. Dec 3, 2009 #1
    For [itex] n \geq m \geq 0[/itex], V is an m dimensional subspace of [itex]\mathbb{R}^n[/itex] and [itex]X=\mathbb{R}^n \backslash V[/itex].

    Let [itex]\pi_0 ( X ) = X / \sim[/itex] be the identification space of X where [itex]x_0 \sim x_1[/itex] if there exists a path [itex]\alpha : [0,1] \rightarrow X[/itex] with [itex]\alpha(0)=x_0, \alpha(1)=x_1[/itex].

    I'm asked to find the no. of path components, [itex] | \pi_0 (X) |[/itex] (the answer varies with m and n).

    i've had a go...

    my understanding so far is that if m=0 then we remove a point from [itex]\mathbb{R}^n[/itex] but we can still create paths from every other point in [itex]\mathbb{R}^n[/itex] to every other point in [itex]\mathbb{R}^n[/itex] so [itex] | \pi_0 (X) | = 1[/itex] provided [itex]n>1[/itex]. if [itex]n=1[/itex] then [itex] |\pi_0 (X)|=2[/itex]. if n=0 then i guess the set of path components would empty as X is empty.

    extrapolating this to higher dimensional m and n,
    surely [itex]|\pi_0 (X)| = \begin{cases} 1 \text{ if } n > m+1 \\ 2 \text{ if } n=m+1 \\ 0 \text{ if } n=m \end{cases}[/itex]

    what do you reckon?
    Last edited: Dec 3, 2009
  2. jcsd
  3. Dec 4, 2009 #2
  4. Dec 5, 2009 #3


    User Avatar
    Science Advisor

    I would recommend that you start by looking at the 2 and 3 dimensional case.

    In R2 (n= 2), V would have to be a line through the origin (m= 1) and X would be the two half planes. We can find a path between any two points in one half plane, but not between two points in different half planes. There are two components.

    In R3 (n= 3), V could be either a line through the origin (m=1) or a plane through the origin (m= 2). In the first case, m= 1, we can find a path between any two points so the number of components is 1. In the second case, m= 2, R3 is divided into two half-spaces and we can find a path between any points in the same half-space but not between points in different half-spaces. The number of components is 2. You should be able to see how this generalizes easily.
  5. Dec 5, 2009 #4
    isn't there also an m=0 case (i.e. a point to consider) in both those examples? however, clearly, we can still connect any two points in [itex]\mathbb{R}^2 \backslash \{ p \}[/itex] or [itex]\mathbb{R}^3 \backslash \{ p \}[/itex] so the number of path components is still 1.

    anyway generalising to higher dimensions, we would getwhat i wrote in my first post would we not?
    also, we have to consider what happens when n=m, doesn't that give 0 copmonenets?
  6. Jan 4, 2010 #5
    I'm doing EXACTLY the same question as this! Saw it hadn't been replied to for a while.
    I think your answer seems right, generalising to higher dimensions from what HallofIvy said does seem to give what you wrote. I agree the m=0 case gives 1 component, not sure about the n=m case though, I'd be grateful for another opinion, or if you explained why you think its 0.
  7. Jan 6, 2010 #6
    hopefully someone can confirm but if n=m then V=R^n and X={} so X has size 0 which means the quotient space X/~ is empty and therefore the size of that space i.e. the number of path components will be 0.

    however, i have only managed to do this by looking at various examples. do you have any idea of how to go about proving this?
  8. Jan 6, 2010 #7
    This thread is pretty old so I assume it's ok to give a pretty much complete answer.

    Choose a basis [itex]b_1,b_2,\ldots,b_m[/itex] of V and extend it with [itex]b_{m+1},\ldots,b_n[/itex] such that [itex]b_1,\ldots,b_n[/itex] is a basis for [itex]\mathbb{R}^n[/itex].

    Let us first consider the case [itex]n > m+1[/itex]. Let x be an arbitrary point of X. We wish to show that x is path-connected to [itex]b_i[/itex] for i>m in X. We can express x in terms of the basis:
    [tex]x = \sum_{i=1}^n x_i b_i[/tex]
    Let [itex]i > m[/itex] be some index such that [itex]x_{i} \not = 0[/itex], and let [itex]j > m[/itex] be some other index than [itex]i[/itex]. Then define the path:
    [tex]f(t) = (1-t)x + tx_{i}b_{i}+tb_{j}[/tex]
    Clearly this is entirely in X since the [itex]i[/itex]'th coordinate is constantly [itex]x_{i}[/itex], and it is a path from x to [itex]x_{i}b_{i}+b_j[/itex]. Now define
    [tex]g(t) = (1-t)(x_{i}b_{i}+b_{j}) + tb_j[/tex]
    The j'th coordinate is constantly 1 so this path is entirely in X, and therefore x is path-connected in X to [itex]b_j[/itex]. Applying this again to [itex]b_j[/itex] we see that x is path-connected in X to any basis-element, and since x was arbitrary so is any other element y and therefore [itex]x \sim y[/itex]. This proves that X has a single path-component.

    Now consider the case [itex]n=m+1[/itex]. A point x is in X if and only if its nth coordinate is non-zero. First let A be the subset of X consisting of all points with nth coordinate positive, and let B be the subset of X consisting of all points with nth coordinate negative. Clearly A and B are disjoint and their union is X. We claim that A and B are the path-components of X. Let
    [tex]x = \sum_{i=1}^n x_i b_i[/tex]
    be an arbitrary point in A. Then define:
    [tex]f(t) = (1-t)x+tb_n[/tex]
    Then the n'th coordinate is [itex](1-t)x_n + t > 0[/itex] so f is a path in X from x to [itex]b_n[/itex]. Thus A is path-connected. In the same way we see that B is path-connected. Now assume that there is a path [itex]p : I \to X[/itex] from [itex]b_n[/itex] to [itex]-b_n[/itex]. Let [itex]\pi_n : \mathbb{R}^n \to \mathbb{R}[/itex] be the projection of the nth coordinate, then let [itex]p' = \pi_n \circ p[/itex]. p'(0) = 1, p'(1) = -1 and p' is continuous so there is some point [itex]t \in (0,1)[/itex] such that p'(t) = 0, but this is a contradiction because then [itex]p(t) \in V[/itex]. Thus A and B are separate path components.

    Finally the case n=m. Then [itex]V = \mathbb{R}^n[/itex] since V is a subspace of [itex]\mathbb{R}^n[/itex] of dimension n. Thus [itex]X = \emptyset[/itex] which means it has no path-components.

    This confirms your original formula.
  9. Jan 6, 2010 #8
    when considering the path f, don't u mean to say the ith coordinate is constantly 1 for i>m and that proves f is definitely in X?
    how do you know the ith coordinate is always 1? we just assumed it was non zero.
    also why do you need to consider the path g?
    why can't you just consider a path such as

    [itex]h(t)=(1-t)x+tb_j[/itex] as the we'd only have to work with one path rather than 2?

  10. Jan 6, 2010 #9
    No I meant that it was constantly [itex]x_i[/itex] as stated. To see this let us expand x and project onto the ith coordinate (let [itex]\pi_i : \mathbb{R}^n \to \mathbb{R}[/itex] be the projection onto the ith coordinate):
    [tex](\pi_i \circ f)(t) = \pi_i((1-t)x + tx_{i}b_{i}+tb_{j})[/tex]
    Now let us consider the ith coordinate for each term. The term (1-t)x has ith coordinate [itex](1-t)x_i[/itex], the term [itex]tx_{i}[/itex] has ith coordinate tx_i and the last term has ith coordinate 0. Thus the total ith coordinate is [itex](1-t)x_i+tx_{i}=x_i[/itex].

    You're right that this is a better choice. I didn't do it because I didn't immediately think of a way to use a single path and avoid having our path intersect V.
  11. Jan 6, 2010 #10
    ok. thanks.
    so we're showing that x is path connected to b_i for i>m but how does this argument rule out the possibility of x being path connected to some point in V? i.e. x being path connected to some b_i with i<=m?

    then in the n=m+1 case, why does x have to have x_n=0 to be in X? what if V was an n-1 dimensional subspace of R^n that consisted of everything except the x_n=0 line?

    and then u take the nth coordinate of f(t)=(1-t)x+tb_n as (1-t)x_n+t why is it not (1-t)x_n+tb_n?
    i think i'm getting confused with notation here, is b_n still just the basis of X? it's nothing to do with B, right?
    and finally, why is [itex]p(t) \in V[/itex]?

    thanks again
  12. Jan 6, 2010 #11
    We are considering the topological space X, and as V is disjoint from X it really makes no sense to talk about a path from X to V in this context. What we're doing is splitting X into path components. To talk about path between X and V we would have to consider all of [itex]\mathbb{R}^n[/itex], but then every point x would be connected to every point y by the path (1-t)x+ty. The important part is that in X we have shown that every point is path-connected to a basis element, which is then connected to all other points of X so by transitivity x is connected to all other points of X. This shows that we have one path component in X.

    For two reasons:
    1) What you described isn't a vector space so it isn't a possibility. For instance the vector 0 isn't in the V you described.
    2) By definition [itex]b_1,b_2,\ldots,b_{n-1}[/itex] is a basis for V and [itex]b_1,\ldots,b_n[/itex] is a basis for [itex]\mathbb{R}^n[/itex]. Thus if we write a point uniquely as:
    [tex]x=\sum_{i=1}^n x_i b_i[/tex]
    Then we know that if x is in V, then it can be written uniquely as a linear combination of the first n-1 vectors so by uniqueness we must have [itex]x_n=0[/itex].

    You confuse the coordinate with the actual basis vector times the coordinate value. For instance if x,y,z is a basis of a 3-dimensional space, then 3x+2y has 3 as it's x-coordinate, 2 as its y-coordinate and 0 as its z-coordinate. It doesn't have 3x as a coordinate. b_n is a basis vector so tb_n is the vector with t as the nth coordinate and 0 as all other coordinates.

    Yes throughout my answer [itex]\{b_i\}[/itex] retains its meaning as the basis.

    Well we showed that the nth coordinate of p(t) is 0 so we can write it uniquely as:
    [tex]x=\sum_{i=1}^{n-1} x_i b_i[/tex]
    and since [itex]b_1,\ldots,b_{n-1}[/itex] is a basis for V this shows that p(t) is in V.
  13. Jan 6, 2010 #12
    a basis element, which is then connected to all other points of X so by transitivity x is connected to all other points of X. This shows that we have one path component in X.[/QUOTE]
    i take it that it's obvious enough that we can just assume that the basis elements are path connected to all other points without proof?

    i don't understand this. we showed the n'th coordinate of [itex]x \in X[/itex] was 0, not the nth coordinate of p(t) was 0.
  14. Jan 6, 2010 #13
    That is what we have shown. In the first part we showed that an arbitrary point x is path-connected to all basis element of index >m except one. If we let [itex]b_j[/itex] be the arbitrary points, then we get that all basis elements are path-connected to each-other so we get a sequence of paths:
    [tex]x \sim b_j \sim b_k \sim y[/tex]

    In this comment I accidentally misused my own notation. I meant,
    [tex]p(t) = \sum_{i=1}^{n-1} x_i b_i[/tex]
    it was the nth coordinate of p(t) we showed was 0 because we considered the function p'(t) which is the projection of p(t) onto its nth coordinate.
    (note that we don't really use the name x for anything at this point)
  15. Jan 6, 2010 #14
    Rasmhop, you say that the path labelled h suggested by latentcorpse is fine. How can you guarantee that this path doesn't cross V? Wasn't that the point of introducing an x_i in your first path g?
  16. Jan 6, 2010 #15
    i think that's also what i'm still struggling to get my head around.

    u said in post 13 that u have a path from x to a basis element and then from that basis element to all other basis elements and then from those other basis elements to any other point in X but in post 7 (where u had your f and g), your map goes from x to x_ib_i+b_j and then from x_ib_i+b_j to b_j. how are these the same thing?
  17. Jan 6, 2010 #16
    For t<1 the ith coordinate of h(t) is [itex](1-t)x_i \not= 0[/itex] (since by assumption j is different from i). Thus for t<1 we have h(t) is in X, and for t=1 we have [itex]h(t)=b_j\in X[/itex]
  18. Jan 6, 2010 #17
    The only conditions we set on i and j was that [itex]x_i \not = 0[/itex] and j is different from i and >m. My composite map went from x to b_j, the intermediate step wasn't essential. Now j could have been any index >m except i so we have a path from x to all basis elements of index >m except b_i. However if we now perform the same process on b_j we see that we get a path from b_j to all other basis elements of index >m including b_i. Thus we can go from x to any basis element. We can also go from y to any basis element, so choose some basis element b_i. Path-connectedness is reflexive so we can go from x to b_i, and from b_i to y where both x and y where arbitrary.
  19. Jan 6, 2010 #18
    Isn't showing that an arbitrary x maps to one basis element enough? Why do you need to show it can map to all the basis elements?

    Edit: is it because you assumed x_i is non-zero and want to generalise to any x in X?
    Last edited: Jan 6, 2010
  20. Jan 7, 2010 #19
    It's because otherwise we can't necessarily path-connect x to every other point. For instance if n=m+2 then I wanted to show that x path-connects to both b_{m+1} and b_{m}, but consider the case where x path-connects to b_{m+1}, b_{m_1} path connects to b_{m+1}, b_m path-connects to b_m and y path-connects to b_m. Then every point we considered is path-connected to a basis element, but we can't path-connect x to y from this information.

    Assuming that x_i is non-zero for some i > m is equivalent to assuming x is in X, so this is already general. This is because otherwise we could express x as a linear combination of b_1,b_2,...,b_m, but then x is in V. If x has some x_i with i >m non-zero then this can't be done so x isn't in V and therefore it is in X.
  21. Jan 7, 2010 #20
    ok. i wrote this out to try and make sense of it and it's all fine except for two things:

    we assumed [itex]x \inX[/itex] iff [itex]x_i \neq 0[/itex] for some [itex]i>m[/itex], how then can we assume that showing the ith coordinate of [itex]h(t)[/itex] to be [itex](1-t)x_i[/itex] tells us [itex]h(t) \in X[/itex]
    is this because [itex]i>m[/itex] is fixed and so we can write [itex]h(t)=\displaystyle \sum_{j=1}^n x_jb_j[/itex] (since we know there is at least one [itex]x_i \neq 0[/itex] for [itex]i>m[/itex])?

    secondly, i figured i better show x is actually paht connected to y
    consider y begin path connected to b_j by [itex]f(t)=(1-t)y+tb_j[/itex]
    this means that [itex](f^{-1} \circ h)(t)[/itex] should be a map from x to y but there are a couple of problems with this. i find [itex]f^{-1}(t)=\frac{t-y}{b_j-y}[/itex] and this gives [itex]f^{-1}(h(t))=\frac{(1-t)x+tb_j-y}{b_j-y}[/itex] but then [itex]f^{-1}(h(0))=\frac{x-y}{b_j-y} \neq x[/itex] and [itex]f^{-1}(h(1))=1 \neq y[/itex]
    can you see where i'm going wrong here?
  22. Jan 7, 2010 #21
    x with coordinates (x1,x2,...,xn) is in V if and only if it's a linear combination of b1,b2,...,bm so x is in V iff xi=0 for all i > m. Or equivalently x is in X iff xi is different from 0 for some i > m. Thus if we can show h(t) has non-zero ith coordinate that tells us that h(t) is in X.

    You seem to mistake function composition with path composition. If we want to compose two paths p and q we don't take their function composite, but rather we define their composite:
    [tex](q*p)(t) = \begin{cases} p(2t) & \qquad t \in [0,1/2] \\ q(2t-2) & \qquad t \in [1/2,1] \end{cases}[/tex]
    which is a path if p(1)=q(0) and it goes from p(0) to q(1). Similarly the inverse path of p is not given by the function inverse, but rather by p(1-t) which is a path from p(1) to p(0).
  23. Jan 7, 2010 #22
    if i wanted to write out the path components, would i be ok to write

    [itex]\pi_0(X)=\begin{cases} \{ A, B \} \qquad n>m+1 \\ \{ X \} \qquad n=m+1 \\ \emptyset \qquad n=m \end{cases}[/itex]

    or should it be [itex]\pi_0(X)= \{ \emptyset \}[/itex] for [itex]n=m[/itex]

  24. Jan 7, 2010 #23
    Your first suggestion is the right one. There are no path components so the set is {}=Ø. Your second suggestion would imply that Ø is a path-component, but by definition path-components are non-empty.
  25. Jan 7, 2010 #24
    ok. i'm now trying to show [itex]\pi_0(X)[/itex] has the discrete topology.

    The topology on [itex]\pi_0(X)[/itex] is [itex]\mathfrak{U}=\{ \emptyset, A,B,X \}[/itex]. Is this true? shouldn't [itex]\pi_0(X) \in \mathfrak{U}[/itex]? Will [itex]\mathfrak{U}[/itex] change according to [itex]n[/itex] and [itex]m[/itex]? I thought about this but it seemed to not work i.e. the topology must be the same all the time.

    so regardless of the value of [itex]n[/itex] and [itex]m[/itex], all the subsets of [itex]\pi_0(X)[/itex] are in [itex]\mathfrak{U}[/itex]. Is it ok to just say that
    [itex]n=m[/itex] we have [itex]\emptyset \subseteq \pi_0(X)[/itex] and [itex]\emptyset \in \mathfrak{U}[/itex]
    then [itex]n=m+1[/itex] we will have [itex]\{ X \} \subseteq \pi_0(X)[/itex] and [itex]X \in \mathfrak{U}[/itex]
    and finally [itex] n>m+1[/itex] we will have [itex]A \subset \pi_0(X)[/itex] and [itex]A \in \mathfrak{U}[/itex]. [itex]B \subset \pi_0(X)[/itex] and [itex]B \in \mathfrak{U}[/itex]. and [itex]A \cup B = X \subseteq \pi_0(X)[/itex] and [itex]X \in \mathfrak{U}[/itex].

    is that ok? i'm definitely on the right lines i think but just not 100% sure i've covered everything.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook