# Orientable Manifold with Boundary

kakarukeys
How does the orientation on M induce an orientation on the boundary of M?

I follow the book Lectures on Differential Geometry by Chern, do not understand the proof.

The proof is
the Jacobian Matrix of the transformation between coordinates of two charts has positive determinant (oriented charts), so the smaller Jacobian Matrix with one row and one column deleted (corresponding to the only one coordinate axis that runs away from the boundary) has positive determinant.

Please do me a favour by explaining the proof clearly, or give me another easier proof.

Staff Emeritus
Gold Member
Hrm.

I would have said "Because it does in Euclidean space", but I don't know how easy it is to prove you can patch together orientations of the pieces of the boundary when it spans multiple charts.

Homework Helper
how does counter clockwise orientation on the upper half plne induce an orientation on the real line?

by noticing that at a point of the real line, one can distinguish between an arrow pointing into the upper half plane and one pointing out of it.

so a vector v along the x-axis is oriented positively if, when it is supplemented by an arrow w pointing into the upper half plane, we get an oriented (counterclockwise) basis <v,w> for the (tangent space to the) upper half plane.

Homework Helper
oh, that is the euliden space acse. but an oriented manifold is by definition one that admits an orientation preserving atlas, so those charts do preserve the euclidean orientation.

kakarukeys
Hrm.

I would have said "Because it does in Euclidean space", but I don't know how easy it is to prove you can patch together orientations of the pieces of the boundary when it spans multiple charts.

Suppose we have charts that cover regions not containing the boundary and charts that cover regions containing the boundary. They are oriented (coordinate transformations among them have positive-determinant Jacobian matrix).

for a region D, let $${u^1, u^2, ..., u^m}$$ be the coordinates, suppose $$u^m = 0$$ is the boundary.

Let $$du^1 \wedge du^2 \wedge ... \wedge du^m$$ be the positive orientation.

then since $$du^1 \wedge du^2 \wedge ... \wedge du^{m-1}$$ is non-zero and gives an orientation on the boundary contained in region D. Minus sign gives another orientation.

if we can transform to other charts on boundary, preserving the orientation, we can extend the orientation to the entire boundary.

And thus we need to show

$$\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0$$
implies
$$\frac{\partial (v^1, ..., v^{m-1})}{\partial (u^1, ..., u^{m-1})} > 0$$

I don't understand the proof of the above in the book, anyone has any idea?
I can type the proof out, if you want.

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Homework Helper
did my comments mean nothing at all?

Homework Helper
another suggestion: just ebcause chern is a great mathematiciazn does not mean he writes learnable boks. read guillemin and pollack instead. or my notes. i will send them to you if you give me an address.

kakarukeys
I am still struggling with the steps of the proof.

kakarukeys
Homework Helper
the point is that if you have an n-1 dimensional subspace spanned by the first n-1 vectors in a basis, and then the last vector points in a giuven direction, if you choose another such absis, the change of basis matrix will be in 2 blocks.

there will be an n-1 by n-1 block in the upper left corner say, and a positive number in the lower right corner, and all zeroes otherwise in the last row and last column.

hence the determinant of the whole matrix being positive forces also the determinant of the upper left block to be positive also.

but i warn you, chern's writings are the worlds most terse and merciless. they are also non conceptual and highly compoutational. he was a great great mathematician but i do not recommend his works to anyone as the sole source for learning any subject at all.

here is a review by another outstanding mathematician:
""This excellent and polished book may not be suitable for the very beginning student, but it is highly recommended for all mathematicians, from the advanced undergraduate student to the experienced professor. For many mathematicians it will be a work of reference for their research. It will be welcome by physicists."

Prof. F Hirzebruch
Max-Planck Institute, Bonn"

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Homework Helper
i will try to pdf my notes and make them sendable. but they are just a rewrite of the treatment in guillemin and pollack which may be in your library.

Homework Helper

kakarukeys
mathwonk said:
i will try to pdf my notes and make them sendable. but they are just a rewrite of the treatment in guillemin and pollack which may be in your library.

Please don't if it is too laborious, I don't want to waste your precious time.

mathwonk said:
the point is that if you have an n-1 dimensional subspace spanned by the first n-1 vectors in a basis, and then the last vector points in a giuven direction, if you choose another such absis, the change of basis matrix will be in 2 blocks.

there will be an n-1 by n-1 block in the upper left corner say, and a positive number in the lower right corner, and all zeroes otherwise in the last row and last column.

hence the determinant of the whole matrix being positive forces also the determinant of the upper left block to be positive also.

In my case, I only have
$$\frac{\partial v^m}{\partial u^m} > 0$$
$$\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0$$

There may not be zeroes in the last row and last column.

Homework Helper
well if you think about the definition of a matrix of a linear transformation, the last row is hte image of the last vector, expanded in terms of the basis ofm the target space.

now we have chsoen both bases to end with a vector pointing into the manifold, so i guess al i can say is that the last entry of the last comloumn is positive, since the component of an inner pointing vector is a positive multiple of another inner pointing vector.

but if you look at the previous columns, you are expanding the images of the basis vectors for the boundaryu space, in terms of anmother absis for that same subspace. so there is no need to use the last inner pointing vector in that expansion. so every columnh except the last has a zero in the last entry.

thus the last row is all zeroes except for the alst right bottom corner. then the same argument applies. i.e. the determinant is positive iff the determinant of the upper left block is positive. how about that?

kakarukeys
your argument is sound unless the following is true.

a cotagent space at a point on the boundary is a direct sum of the cotagent space of that point when the boundary is considered an independent manifold and its complement.

Every first m - 1 basis vectors given by any chart span the first subspace, and the last vector pointing into the manifold always spans the other subspace.

I think it's true by definition of the chart. or is it true?

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kakarukeys
only then can it be that

when we transform from u basis to v basis,
the first m-1 vectors are shuffled.
and the last vector is only multiplied by a constant.

kakarukeys
Ok, forget about the cotangent space direct sum.
Here's the argument

We know given two charts, at a point on boundary
the first m - 1 cotangent vectors obtained $$\{du^1, ... du^{m-1}\}, \{dv^1, ... dv^{m-1}\}$$
span the cotangent space of the boundary at the point, by definition of the chart.

so entries of last column of the m by m Jacobian Matrix we considered are all zeros except the last one. (Because $$\{dv^1, ... dv^{m-1}\}$$ can be expressed solely as a linear combination of $$\{du^1, ... du^{m-1}\}$$).
And so the conclusion follows.

P/S the last vector $$dv^m$$ need not be multiple of $$du^m$$, it can contains some linear combination of $$\{du^1, ... du^{m-1}\}$$.

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Homework Helper
it depends on your definition of the complementary space. some people take the normal space to be the quotient of the full tangent space by the tangent space to the boundary. when you do that lkinear combinations of baoiundary tangent vectors are set to zero, so the normal vectors are scalar multiples of each other in that quotient space. that is the point of view in guillemin pollack, and hence also in the pdf notes i sent you.

but you seem to understand it now.

kakarukeys
I thought I understand, but when I tried a simple example, I couldn't get what I want.

Consider the Manifold, closed interval [0, 1]

Using two charts:
[0, 1) -> [0, 1) $$\phi_1(x) = x$$

(0, 1] -> [0, 1) $$\phi_2(x) = -x + 1$$

The first chart is positively oriented, the second chart is negatively oriented.

So the orientation of the boundary point is +1 and -1 respectively.

since $$+1 \wedge dx = dx =$$ gives the positive orientation on the first chart.

since $$-1 \wedge dx = -dx$$ gives the "positive orientation" on the second chart.

then applying Stokes' theorem we have,

$$\int^1_0 df = \int_{0, 1} f(x) = f(0) - f(1)$$

since the 2nd chart is negatively oriented.

Doodle Bob
kakarukeys said:
I thought I understand, but when I tried a simple example, I couldn't get what I want.

Consider the Manifold, closed interval [0, 1]

Using two charts:
[0, 1) -> [0, 1) $$\phi_1(x) = x$$

(0, 1] -> [0, 1) $$\phi_2(x) = -x + 1$$

The first chart is positively oriented, the second chart is negatively oriented.

The above example above doesn't count since you've chosen an atlas that does not induce an orientation on the interior manifold. Orientability implies there exists an atlas with coordinate charts whose Jacobians have pos. determinant, but it doesn't mean that any atlas of the manifold will have that property -- as you can see above. In order to use Stoke's theorem you need to settle on a particular orientation (you do after all have a choice of two) and use that.

kakarukeys
The above example above doesn't count since you've chosen an atlas that does not induce an orientation on the interior manifold. Orientability implies there exists an atlas with coordinate charts whose Jacobians have pos. determinant, but it doesn't mean that any atlas of the manifold will have that property -- as you can see above. In order to use Stoke's theorem you need to settle on a particular orientation (you do after all have a choice of two) and use that.

I am well aware of that. So I put a minus sign in front, whenever I want to compute integral using the chart with negative orientation. The effect is same as sticking to a particular atlas with oriented charts. So there is nothing wrong in my example, but the answer is wrong.

Doodle Bob
kakarukeys said:
I am well aware of that. So I put a minus sign in front, whenever I want to compute integral using the chart with negative orientation. The effect is same as sticking to a particular atlas with oriented charts. So there is nothing wrong in my example, but the answer is wrong.

I think you're confusing which determinants of what mappings need to be positive or negative in order to induce an orientation. An atlas with coordinate charts induces an orientation if and only if the Jacobians of the transitions functions, i.e. the $$\phi_1 \circ \phi_2^{-1}$$ on the intersection of open sets of the atlas, all have *positive* determinant.

If one of those transition functions has a neg. det. Jacobian, then that means you've switched from one orientation to the other on that intersection, just as you've done in your example.

The two possible orientations on [0,1] are given by the volume forms dx and -dx. phi_1 induces dx, whereas phi_2 induces -dx. Thus, the mapping $$\phi_1 \circ \phi_2^{-1}$$ is in fact orientation-reversing.

kakarukeys

I haven't been able to find an atlas which has oriented charts that cover [0, 1]
do you know one?

Doodle Bob
kakarukeys said:

I haven't been able to find an atlas which has oriented charts that cover [0, 1]
do you know one?

just change phi_2 to phi_2(x)=x+1. Trivial, yes, but [0,1] is a rather trivial manifold.

kakarukeys
In that case, you are also using a chart that does not satisfy the requirements given in the definition.

that $$\phi_2 : U \rightarrow H^n$$
$$H^n = \{x : x^n \ge 0\}$$
boundary is given by $$\{x : x^n = 0\}$$

I am not sure if you use such chart, the stokes' theorem still holds.

anyway please show me how you prove fundamental theorem of calculus from stokes' theorem.

Homework Helper
well you are asking the world's most difficult question, getting minus signs correct. anyway the case of a one dimensional manifold is a little special, because orientation is usually defined by choosing an ordered basis of the vector space.

now if the space has positive dimension then every ordered basis can be changed by replacing the first vector by minus itself to get the opposite orientation. the boundary of a one dimensional manifold however has dimesnino zero, and there is only one basis of that space, the empty basis.

so in that one case the definition of orientation is different, just the choice of +1 or -1.

now in the usual case we define orientation of a boundary by saying, take any basis for the boundary such that preceding it by an outward pointing normal vector gives an oriented basis of the manifold.

now if we precede the empty set by an outward pointing normal vector, we do get an oriented basis for the manifold [0,1] at the point 1 but not at the point 0.

so perhaps we should say that preceding by the number +1 means keeping the orientation and preceding by the number -1 means changing it.

then we would get that the correct orientation on the boundary of [0,1] is +1 at 1, and -1 at 0, since the outward normal is positively oriented at 1 but not at 0.

notice the arguments about determinants of matrices make no sense in this case, and assume that the dimension of the boundary is at least one.

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