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: General and specific existence and uniqueness proofs

  1. Jan 14, 2006 #1
    Short question: Can anyone provide me with a nice synopsis of how to go about proving the "existence" of some object as often requested in math questions such as, "prove that X really exists and is unique"? In other owrds, in general, when presented with an "existence" question, is there a nice way to interpret such a question, and set about trying to answer it?

    Long form of question:
    I can handle the uniqueness part usually (or at least I know what I need to prove). And for many problems existence is fairly simple to ascertain, but I got stuck for example trying to understand how to demonstrate existence of a pullback map of a 1-form. This bugged me because it exposed to my mind a general lack of education about how to attack general "existence" type questions---i.e., when it's not obvious that you have an equation with an unknown and a putative solution that can be just plugged in to the original equation. When you have an existence question that is not in the form of an equation in an unknown and a conjectured solution, then how do you go about establishing existence?

    Specific example:
    Taken from "Gauge Fields, Knots and Gravity", Baez and Munian.
    (Exercise 32.): Define a pullback of a 1-form $(\phi^\ast $ by

    \[ \label{eq1}
    (\phi^\ast\omega)_p = \phi^\ast (\omega_q)

    where $q=\phi(p)$, and $\omega_q$ is an element of the cotangent space on a manifold $N$ say and $\phi : M \rightarrow N$ is a map.

    Now we are given that for a tangent space vector $v$ from $T_pM$, and $\nu$ from $T_q N$ a cotangent vector, then

    \[ (\phi^\ast \nu) (v) = \nu(\phi^\ast v) \]

    again $q=\phi(p)$. (So $\p\in M$ is mapped by $\phi$ to $q\in N$.)

    So locally we have a pullback of a cotangent vector. The Exercise is to prove that the former expression (\ref{eq1}) for 1-forms is a global extension of the local pullback. Thus,

    "Exercise 32. Show that the pullback of 1-forms defined by the above formula (\ref{eq1}) really exists and is unique."

    I came up with an argument to answer this question, but I cannot fathom if it is correct or is exactly what is intended by an "existence" proof.

    I somewhat naively just argued that at any $p\in M$ the map $\phi^\ast\omega_q$ can act on a tangent vector $v_p$ (so $v\in T_pM$) by the local pullback to $\omega(\phi^\ast v_p)$. But $v_p$ could be a local value of a vector field $v$ at the point $p$. It's pullback $\phi^\ast v$ is then by definition a vectr field on $N$, in other words the pullback of the tangent vecotr $v_p$ is a tangent vector on $N$. This, I believe, establishes that the formula

    \[ (\phi^\ast\omega)_p = \phi^\ast (\omega_q) \]

    is well-defined, i.e., it "exists"! Or have I missed something in this justification?

    My problem is this: I cannot tell if I have really demonstrated existence of the pullback of 1-forms. The only other existence theorems I've come across in physics are usually those involving solutions to equations defined in terms of unknowns (or DE's where a solving function is given and one can check it exists wherever it is supposed to), whereas above we have what looks more like an equality or definition, not an equation in search of a solution variable.

    But after completing my agument I seemed none the wiser about what I accomplished. I thus thought I'd write to physics forums here to see if anyone had a general perspective to offer on how to interpret and set about demonstrating mathematical existence type questions.

    ASIDE: As for uniqueness of the pullnack for 1-forms, supposing two pullbacks of the same 1-form $\omega$ on $N$ say, $phi^\ast \omega$ and $\psi^\ast \omega$ yield the same 1-form on $M$, then I can show that the two associated maps $\phi$ and $\psi$ are equal, thus I conclude that the pullbacks are equivalent, which proves uniqueness.
  2. jcsd
  3. Jan 14, 2006 #2


    User Avatar
    Homework Helper

    Existence proofs can be accomplished by demonstating the existence of a thing directly or in directly (e.g.:

    Direct: there exisits a value of x such that x+3=0; proof: x=-3 is such a value.

    Indirect: there exists a value of x such that x+3=0; proof: Let f(x)=x+3. Then f is seen to be continuous (as it is a ploynomial) and we have f(1)=4 and f(-5)=-2, so by the intermediate value theorem, there exists a value of x such that -5 < x < 1, and such that f(x)=0.)

    As for Uniqueness proofs, proof by contradiction is rather handy: having shown the existence of some thing, say x such *blah* is true, then suppose that there exists two such things, say x and x' such *blah* is true, and then show that if *blah* is true for both x and x', then invariably it follows that x=x'.
  4. Jan 15, 2006 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    No there is no universal method of proof. You just have to find a reason why something must exist, be it by construction or an appeal to something non-constructive.

    And 'well-defined' is not the same as 'exists', really, though it might be though part of the proof that what 'exists' is, erm, well, 'well-defined', which usually means 'is independent of any choices made'.

    You also have uniqueness wrong. You must show that the thing you pull back to is unique, not the map \phi which is after all fixed. Ie given any choices (local data) the answer you get is always the same, so this is the well defined part. That is if I give you the data at a point int two different ways then phi^* does the same thing to both.

    Here's an example.

    Show that every real positive number has a well defined positive square root.

    Existence: consider x^2-s for s positive, this is negative at x=0, and positive for x sufficiently large, hence it has a zero at x=r and r^2=s and r is positive. Now we need to show uniqueness: if r and t are square roots, then r^2=t^2 ie (r-t)(r+2)=0 thus r=-t so one is negative, or r=t and we are done.
    Last edited: Jan 15, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook