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Exercise from Guillemin and Pollack's book:

Assume $X \pitchfork Z$, both compact and oriented, and prove directly from the definition that $$I(X,Z)=(-1)^{(\dim X) (\dim Z)}I(Z,X).$$ (I assume $X,Z$ are (sub)manifolds of a manifold $Y\subset R^N$.)

In this case $I(X,Z)$ is the (signed) number of points in $X\cap Z$, where a point $y$ is included with a plus sign if the orientation of $X$ and $Z$ (in that order!) "add up" at $y$ to the orientation of $Y$; otherwise $y$ is counted with a minus sign.

I can see that this holds from the picture in the particular case when $X,Z$ are two "independent" loops on the torus, but I don't know how to generalize this picture to higher dimensions and write a rigorous proof

amWhy
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user557
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    This is really just linear algebra in the tangent space. – Mariano Suárez-Álvarez Mar 17 '18 at 21:52
  • Write out the definition of the orientation of a vector space. – anomaly Mar 17 '18 at 22:03
  • @MarianoSuárez-Álvarez Of course it is, but I have a hard time understanding how to write this proof. So I was hoping at least for some sketch to help me understand how to get started. – user557 Mar 18 '18 at 01:41
  • @anomaly I can write it out, but I don't know how exactly to use it in this problem. (It is a "decision to affix a positive sign to the elements of one (of the two) equivalence class and a negative sign to the others" (two ordered bases are declared to be equivalent if the determinant of the basechange matrix is positive). – user557 Mar 18 '18 at 01:44

1 Answers1

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Suppose $V$ is an oriented real vector space of dimension $n$, with positive ordered basis $B=(v_1,\dots,v_n)$. Let moreover $U$ and $W$ be two subspaces of $V$ such that $V=U\oplus W$, and suppose they are oriented, with positive ordered bases $B_U=(u_1,\dots,u_r)$ and $B_W=(w_1,\dots,w_s)$. Of course, we have $n=r+s$.

We write $B_UB_W$ the concatenation of $B_U$ and $B_W$, that is, the sequence $(u_1,\dots,u_r,w_1,\dots,w_s)$; this is clearly an ordered basis of $V$. We say that the decomposition $U\oplus W$ of the oriented vector space $V$ as a direct sum of two oriented subspaces is positive or negative if $B_UB_W$ is a positive or a negative ordered basis of $V$, respectively, that is, if the change of basis matrix $C(B,B_UB_W)$ has positive or negative determinant, respectively.

Now $C(B_UB_W,B_WB_U)=(-1)^{rs}$ and $C(B,B_UB_W)=C(B_UB_W,B_WB_U)C(B,B_WB_U)$. This implies a once that the "sign" of the decomposition $U\oplus W$ of $V$ is $(-1)^{rs}$ times the "sign" of the decomposition $W\oplus U$.

amWhy
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Mariano Suárez-Álvarez
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  • Thanks for your answer! Could you clarify what is $C_U$? Also, I'm not sure whether I understand how exactly the conclusion in the last sentence implies the result. – user557 Mar 18 '18 at 16:08
  • $C_U$ is just typo for $B_U$. – Mariano Suárez-Álvarez Mar 18 '18 at 18:20
  • To see why this answers the question, conider how you compute the sign of a point of interesection when computing $I(X,Z)$. – Mariano Suárez-Álvarez Mar 18 '18 at 18:22
  • I fixed a couple of typos (at least I believe they were typos), but I still cannot understand how exactly this is related to the original problem. Intuitively, replacing $I(X,Z)$ by $I(Z,X)$ makes me to "flip" $rs$ orientations, but this idea remains vague in my mind. – user557 Mar 20 '18 at 20:32
  • Well, can you tell me exactly how you assign the sign to a point $y$ of the intertsection $Z\cap X$? That is, what exactly does the phrase «the orientation of $X$ and $Z$ (in that order!) "add up" at y to the orientation of $Y$» that appears in your question mean? – Mariano Suárez-Álvarez Mar 20 '18 at 21:11
  • There is a fixed orientation on the ambient space. Consider some intersection point. The tangent space at that point to $X$ has some fixed basis, and so does the tangent space to $Z$. The direct sum of the tangent spaces is the tangent space to the ambient space, so if one appends the two bases, one gets the basis of the ambient space. If the change of basis matrix from the appended basis to the fixed one has positive determinant, we assign $+1$, otherwise assign $-1$. But there are many points at which we need to do the above, and I'm not sure how to collect all this information. – user557 Mar 20 '18 at 21:32
  • Also, in your answer, shouldn't the LHS of the equation whose RHS equals $(-1)^{rs}$ stand for the determinant (as opposed to a matrix)? – user557 Mar 20 '18 at 21:34
  • Do it at each point separately. Honestly, I don't know how you cannot see the connection between that which you describe and what I wrote. – Mariano Suárez-Álvarez Mar 20 '18 at 23:44