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Let M be a smooth manifold. We say that M is orientable if and only if there exists an atlas $A = \{(U_{\alpha}, \phi_{\alpha})\}$ such that for all $\alpha, \beta$, $\textrm{det}(J(\phi_{\alpha} \circ \phi_{\beta}^{-1})) > 0$ (where defined). I'm struggling to understand the reason this definition is made. My question is: What is the intuitive reason for this definition of an orientable manifold?

Thank you!

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I think this is a great question. See my comments to Andres' nice answer to your previous question. I will definitely take a crack at it later on if no one beats me to it. –  Pete L. Clark Dec 27 '10 at 2:07
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I recommend thinking about surfaces. Check and understand that the sphere and torus are orientable, but the projective plane and Klein bottle are not. Check that in these examples the definition agrees with more intuitive notions of orientability (namely "handedness is well-defined"). –  Noah Snyder Dec 27 '10 at 3:16
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Rich: To integrate a differential form on a manifold, you want the manifold to be oriented because the change of variables formula for multiple integrals in R^n requires the absolute value of the Jacobian determinant, whereas when you write a differential form in two different coordinate systems you get a Jacobian determinant with no absolute value sign. So as long as you only work with coordinate systems linked by positive Jacobians, you don't meet absolute value signs in multiple integrals and thus integration of differential forms on the manifold can be defined (with partitions of unity). –  KCd Dec 27 '10 at 13:13
    
Richard: as an entertaining "concept check", apply the above definition of orientability to some $1$-dimensional manifolds with boundary. You'll see that the above definition breaks down and gives you some "non-orientable 1-manifolds". It turns out to only fail for 1-manifolds with boundary, i.e. in all other dimensions the above definition agrees with the standard ones. –  Ryan Budney Dec 31 '10 at 18:12
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5 Answers

up vote 6 down vote accepted

I just want to flesh out something Ronaldo said

Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go "continuously" from one to another.

To this end, consider the vector space $\mathbb{R}^n$. Let $Gl = Gl_n(\mathbb{R})$ be the set of all invertible linear maps from $\mathbb{R}^n$ to itself and let $M = M_n(\mathbb{R})$ be the set of all linear maps (possibly noninvertible) from $\mathbb{R}^n$ to itself. The space $M$ is naturally a vector space and is naturally isomorphic to $\mathbb{R}^{n^2}$. Use this identification to topologize $M$. The important thing about this topology is that polynomials in the entries of a matrix are continuous - this is all we'll use about it.

We have a map (the determinant) $det:M\rightarrow \mathbb{R}$, which is a continuous homomorphism (it's given as a polynomial in the entries of the matrix).

We can identify $Gl$ with $det^{-1}(\mathbb{R}-\{0\})$, i.e., $Gl$ is an open subset of $M = \mathbb{R}^{n^2}$. Give $Gl$ the subspace topology.

Now we get to the main claim: With this topology, $Gl$ has precisely two path components. I'll prove this in a second. In the mean time, notice that this makes what Ronaldo said precise: If we start in the component of $Gl$ containing the identity matrix, then a path of elements in $Gl$, starting at the identity matrix, gives rise to a path of basis vectors starting with whatever choice of basis you've started with - it gives you a way to continuously go from one basis to another. But you'll never be able to get to things of opposite orientation because that would require a path from one component of $Gl$ to another, which is impossible.

Now I'll establish the claim that $Gl$ has precisely two components. First notice that is must have at least two components because $det:Gl\rightarrow \mathbb{R}^n-\{0\}$ is continuous, surjective, and the range has two components. Let $X^+\subseteq Gl$ be those matrices of positive determinant and $X^-$ those of negative determinant, so we have $Gl = X^+\cup X^-$.

Next, I claim that $X^+$ and $X^-$ are homeomorphic. To see this, let $A\in X^-$ be anything at all and consider the map $f:X^+\rightarrow X^-$ given by $f(B) = BA$. Since matrix multiplication is given by polynomially in the entries of the matrices, $f$ is continuous. It has inverse $f^{-1}(B) = BA^{-1}$, which is continuous for the same reason $f$ is. Thus, $f$ is a homeomorphism.

All the following maps and paths will be continuous for the same reason: they'll be given as polynomials in the entries of the matrices involved.

Finally, I just need to show $X^+$ is path connected. So, let $A\in X^+$ be arbitrary. I want to connect it to the identity matrix by a path. Let $E_{ij}$ denote the matrix of all 0s with a 1 in the $ij$th slot. Let $F_{ij}(t) = I + tE_{ij}$. Notice that for $i\neq j$, $F_{ij}$ is triangular, so has determinant 1. Left multiplying by $F_{ij}(t)$ corresponds to the row operation of adding $t$ times the $j$th row to the $i$th row, which doesn't change the determinant if $i\neq j$.

Since the $F_{ij}(t_0)$ are just row operations, we know we can multiply by several of them (each with a different appropriate $t_0$) to bring $A$ to a diagonal form without changing the determinant. By multiplying by $F_{ij}(t)$ and letting $t$ go from $0$ to $t_0$, we make a path out of this. Doing this for all the $F_{ij}(t_0)$ simultaneously gives us a path from $A$ to a diagonal matrix where for all time, the determinant is constant.

We've now connected $A$ to a diagonal matrix $D$ by a path in $X^+$. It remains to connect the diagonal matrix to the identity. We know all the entries of $D$ are nonzero because $D$ has positive determinant. If an entry $d$ of $D$ is positive, use the path $t + (1-t)d$ in that slot to continuously change $d$ to $1$, all the while keeping the determinant nonzero. If $d$ is negative, using the same trick, we can change $d$ to $-1$. At this point, the determinant must be $1$ - it's positive because we're in $X^+$ and it's a product of $\pm 1$s.

We've now reduced $D$ to a new diagonal matrix $E$ whose entries are plus and minus 1s. If all are positive, we're done, so assume we have a $-1$ somewhere. Since $E$ is in $X^+$, there must be another entry which is $-1$. For notation, let's assume one $-1$ is in the $i$th position and the other is in the $j$th position. Think of the two $-1$s as $cos(t\pi)$ and think of the $ij$th entry and $ji$th entries as $sin(t\pi)$, all with $t=1$. Now, let t go from 1 to 0. One can easily check the determinant stays $1$, and this gives a path which changes the two $-1$s to $1$s. Doing this finitely many times eliminates all the $-1$s and changes $E$ to the identity matrix.

Thus, we've connected $A$ to the identity matrix by a path, so $X^+$ is connected.

(Kudos to anyone who reads the whole thing - it ended up being WAY longer than I anticipated, but it was a good exercise for me to go through).

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Here's another argument, using Gram--Schmidt instead of elementary matrices: math.stanford.edu/~conrad/diffgeomPage/handouts/gramconnd.pdf –  KCd Dec 27 '10 at 13:03
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thanks for the precise discussion of the subject!+1 –  Ronaldo Dec 27 '10 at 14:03
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I think it is very hard to get a better answer than any of the above. But there is a perspective that has not been mentioned, at least not explicitly. First, I would like to say that I happen to agree with Marianos answer.

The body of your question is much more explicit than your title, so I will address the question in your title since it seems others have addressed the one in the body of your question. I like to think of orientations, or in fact most structures on a manifold, in terms of the tangent bundle on the manifold in question. The tangent bundle on a compact n dimensional manifold $M$ is classified (up to isomorphism) by a homotopy class of map $M \to BO(n)$. If we can lift this map over the map $BSO(n) \to BO(n)$ then the bundle is orientable, and hence so is the manifold. Here, the notion of orientability, our structure is really a fact about the tangent bundle. We can talk about almost complex structures or spin structures in a similar way. The map $BSO(n) \to BO(n)$ is induced by the inclusion $SO(n) \to O(n)$.

Another way is to look at $\Lambda^n TM$, the top exterior power of the tangent bundle of M. Then an orientation is a nowhere zero continuous section of this bundle. Similarly, you can think of an orientation as a choice of generator of the top dimensional integral homology or cohomology. These are all probably less helpful than the above answers, but these are the ways I like to think about what an orientation is.

please let me know if you want me to expand on some of these.

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I really like this explanation. Knowing that an orientation exists iff a certain bundle has a non-zero section matches one's intuition about an orientation giving some kind of "direction" at each point (a determinant) and it helps elucidate the concept of an obstruction cohomology class which is dual to the sometimes unavoidable 0-set. And as you mention, this way of thinking naturally extends to other geometric structures on manifolds. So cool. –  Obidiah Dec 28 '10 at 4:05
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Suppose the universe has shape $M$. Then $M$ is orientable if and only if you can write yourself a note, then begin a journey, and come back to always find that you can still read the note. If $M$ is not orientable, it is possible to return from a journey to find that the note is written backwards and, indeed, that everything around you is written backwards, your friends' faces are all reversed, etc. More seriously, the proteins in your body would have the wrong chirality and I think this would cause problems involving digestion.

It is not known, to the best of my knowledge, whether our universe is orientable. If you are interested in this more intuitive approach to topology and geometry before attacking its rigorous foundations, I recommend reading Weeks' The Shape of Space and being very diligent about doing the exercises.

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"...the proteins in your body would have the wrong chirality and I think this would cause problems involving digestion." - to give this more precision, the problem is that the enzymes of the body are designed to accommodate molecules with a particular "handedness" (what is technically called chirality ), as they themselves possess handedness (much like one cannot wear a right glove on his left hand); if you cannot acquire the amino acids and carbohydrates with the required chirality, you starve. –  J. M. Dec 27 '10 at 5:37
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"... it is not known whether our universe is orientable..." It is actually quite a bit worse than that. Not only do we not know, a lot of Theorems in theoretical cosmology and general relativity presumes orientability. Locally it is of course not an issue, but globally speaking... (One can however give a meta-argument showing that this shouldn't be a problem for deterministic laws of physics operating in an expanding universe.) –  Willie Wong Dec 27 '10 at 13:26
    
Regarding chirality of proteins - check out "Doorways in the Sand" by Roger Zelazny. –  Sam Nead Dec 27 '10 at 15:28
    
What I like about this answer, is it is completely internal knowledge to the universe. –  BBischof Dec 27 '10 at 19:32
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I am of the opinion that overthinking about what things mean at the time one is first learning them can often be counterproductive. I like to quote, and fully agree with, D'Alambert, who famously said

Allez en avant, et la foi vous viendra.

("push on and faith will catch up with you", according to a translation I just googled for)

As for your concrete question: playing tetris, playing with your right hand and linear algebra should have convinced you that a real vector space has two orientations. Now, a manifold provides, among other things, a family of vector spaces with a piece of extra information: a way to move from one of them to others "continuously". We say that the manifold is orientable if we can pick an orientation in each of them in a continuous way.

I promise you that in not a long time, it'll make way more sense and you'll even find it difficult to explain it to others because it'll seem intuitive to you!

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"it'll make way more sense and you'll even find it difficult to explain it to others because it'll seem intuitive to you!" - I liked this part. :D –  J. M. Dec 27 '10 at 2:36
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@Mariano: with all respect, I think this is not a very useful answer. –  Pete L. Clark Dec 27 '10 at 2:36
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Questions about "why is X something" and "what does X mean" are very effective, even essential, ways to learn mathematics at a deep level. Although they may not have "objective" solutions, I've noticed they elicit some of the most inspiring and valuable answers. Some believe we can only make intuitive progress in private, but I strongly disagree. I find the dismissal of these questions a little like hazing in a fraternity, an "I had trouble getting some insights, so you should too" mentality. Let's share the insights we already have so younger mathematicians can develop better ones. –  Obidiah Dec 27 '10 at 6:18
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Furthermore, I believe new tools like stackexchange and MO can be used to distribute "ways of thinking" like never before. Other mathematicians can now share how they think about concepts, how they get to solutions and not just the results. For instance, on MO it has been awe inspiring for me to see how Bill Thurston thinks about math. He goes way beyond objective answers and explains how he visualizes problems. Instead of glorifying intuitive understanding which you can't explain, we should all struggle harder to articulate those valuable intuitions. –  Obidiah Dec 27 '10 at 6:23
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@josephgeorge4: I agree with everything you said. But I think that understanding comes from knowledge, from familiarity, and I am of a position quite distant from that of glorifying intuition: knowledge and familiarity appear to me as utterly necessary conditions for the development of what we call intuition (and surely not sufficient) Surely, I cannot explain Ramanujan, or Thurston. –  Mariano Suárez-Alvarez Dec 27 '10 at 15:04
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Get two intersecting charts $(U,\phi)=(U, (x^i))$, $(V,\psi)=(V, (y^i))$, with $p\in U\cap V$. Then there are two basis for the tangent space of $M$ at $p$: $\{ \frac{\partial}{\partial x^\alpha}\}$ and $\{\frac{\partial}{\partial y^\beta}\}$ . Intuitively, these two bases have the same orientation (in the vector space $T_pM$) if it is possible to go "continuously" from one to another. More precisely, they have $the$ $same$ orientation if the determinant of the matrix of change of basis from $\{ \frac{\partial}{\partial x^\alpha}\}$ to $\{\frac{\partial}{\partial y^\beta}\}$ is positive. If this determinant is negative, we will have to change the direction of one or more axes.

Ok. Given that, the question is: "How to extend this concept of orientations from one tangent space to the entire manifold"? That is where we define the $orientability$ of the manifold: a manifold will be $orientable$ if it admits an atlas such that, given two charts in this atlas with a nonempty intersection, at every point of this intersection the determinant of the matrix of change of basis from one coordinate basis to the other is positive (just like above). That is, the coordinate basis $\{ \frac{\partial}{\partial x^\alpha}\}$, $\{\frac{\partial}{\partial y^\beta}\}$ at a given point generated by any two charts of this "orientable atlas" have all the same orientation.

The matrix of change of basis between two coordinate basis is exactly the jacobian of the coordinate transformation.

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