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From: bacle on 29 Jul 2010 13:32
Hi, everyone:
This is a followup from a while back. W.Dale Hall
was pretty helpful.

If I remember well, he had suggested I use the
preferred orientation on the submanifolds (i.e.,
given a complex basis {w_1,..,w_n}, we get a real
basis {w_1,iw_1,..,w_n,iw_n } . The structure group
for these "induced" bases is Gl(n,R)^+ , which shows
that complex manifolds are orientable.

Still, given the intersection of (not necessarily different) submanifolds X,X' ,I don't see how using the preferred/induced bases on submanifolds helps us construct a positively-oriented basis for T_xX(+)T_xX'
at points of intersection.

Thanks.
From: W. Dale Hall on 31 Jul 2010 14:21
bacle wrote:
> Hi, everyone: This is a followup from a while back. W.Dale Hall was
> pretty helpful.
>
> If I remember well, he had suggested I use the preferred orientation
> on the submanifolds (i.e., given a complex basis {w_1,..,w_n}, we get
> a real basis {w_1,iw_1,..,w_n,iw_n } . The structure group for these
> "induced" bases is Gl(n,R)^+ , which shows that complex manifolds are
> orientable.
>
> Still, given the intersection of (not necessarily different)
> submanifolds X,X' ,I don't see how using the preferred/induced bases
> on submanifolds helps us construct a positively-oriented basis for
> T_xX(+)T_xX' at points of intersection.
>
> Thanks.

I guess I should pitch in, since I seem to have put this line of
argument forward to begin with.

I'll assume we're talking about either an immersed manifold X (that's
how you'd get self-intersections and still be able to think of the
tangent spaces as being n-dimensional), or two embedded submanifolds
X1 and X2 (my later notation for points being the cause for my re-
wording of your original notation for the X's. I'll also assume we're
only talking about points x where the preimage of x consists of two
points (which I'll call x1 and x2, with xi being in Xi). I'll denote
the embedding (respectively, immersion) by f1,f2 (respectively f),
and the target space by Y

Now, if there are two spaces, X1 != X2, we have two possibilities:
the intersection comprises a set of isolated points (which poses
no problem, since you can simply take the orientation induced by
taking the basis for Tf1 (T_x1 X1) followed by that for Tf2 (T_x2 X2)
where Tfi denotes the mapping TXi ---> TY induced by fi. The condition
that the intersection points be isolated prevents any disagreement
among the various orientations at the points of intersection.

If the set of intersection points is not discrete, the only possible
difficulty lies in moving the intersection point x along a closed
curve and having the orientation described above somehow result in
a reversal of orientation. Note, however, that if f1 and f2 are
embeddings, the orientations of Tf1(TX1) and Tf2(TX2) are individually
preserved, and maintaining the ordering "Tf1(TX1) followed by Tf2(TX2)"
similarly preserves orientation.

In the case of an immersion, we have x1 & x2 mapped by the immersion f
to the point x, and the above situation applies again (namely, is the
intersection set discrete or not). There is one additional wrinkle that
can arise in the second (non-discrete) case: taking a closed curve in
the intersection set in Y that starts & ends at x, for which the
preimage gives two paths: p12, starting at x1 and ending at x2, and
p21, starting at x2 and ending at x1.

In this case, you get (at first) the orientation described by taking
Tf(T_x1 X) followed by Tf(T_x2 X), but (at the end) the orientation
described by taking Tf(T_x2 X) followed by Tf(T_x1 X). Let's look at
this in a bit finer detail:

We have basis vectors (over C) for T_x1 X: {u1 ... un}, and similarly
basis vectors (over C) for T_x2 X: {v1 ... vn}. The trick I mentioned
earlier to produce real bases similarly applies:

R-basis for T_x1 X:

{u1 iu1 u2 iu2 ... un iun}

R-basis for T_x2 X:

{v1 iv1 v2 iv2 ... vn ivn}

and we have two orientations for T_x Y:

{u1 iu1 ... un iun v1 iv1 ... vn ivn}

and

{v1 iv1 ... vn ivn u1 iu1 ... un iun}

Note that the mapping of the first to the second looks like this:

[ ]
[ 0(2n x 2n) I(2n x 2n) ] I(k x k) - identity matrix
[ ] 0(k x k) - zero matrix
[ I(2n x 2n) 0(2n x 2n) ]
[ ]

It's a simple matter to verify that the determinant of this
matrix is 1, since it takes an even number (4n, if I've
counted correctly) of column transpositions to turn this
matrix into I(4n x 4n).

I think that does it.

Dale
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