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I want to show that $I(Z,X) = -I(X,Z)$.

So clearly I have two orientations for $X$ and $Z$ each. Do I discuss these four cases, each consider $I(Z,X)$ and $I(X,Z)$? Is this the correct approach and is there a less brute-force way to do so?

Thank you~

enter image description here

1LiterTears
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    You should tell us whet $X,Z,I$ are. I assume $X,Z$ are transverse one-dimensional submanifolds of an oriented two-dimensional manifold and $I$ theri intersection number. In that case, note that the orientaion of $(a,b)$ is the negative of the orientation of $(b,a)$ if $a,b$ form a basis of the tangential space. – Hagen von Eitzen Jul 21 '13 at 22:10
  • Oh @HagenvonEitzen I thought I only need to prove the case in the picture... – 1LiterTears Jul 21 '13 at 22:12
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    I think it's much more fun to guess what the OP could mean; context is so over-rated. – Jesse Madnick Jul 21 '13 at 22:12
  • .....nonono, Mariano(a moderator constantly warning me about this issue) wouldn't see this!! – 1LiterTears Jul 21 '13 at 22:13
  • The assymetry in general will be discussed later, so I assume I only need to prove the transversed two 1-dimensional space as illustrated in the graph. Do I still expect to specify what is X,Z,I or anything else? And thanks for the good laugh, @JesseMadnick. – 1LiterTears Jul 21 '13 at 22:23

1 Answers1

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In general, if $\dim X=k$ and $\dim Z=\ell$, then $I(Z,X)=(-1)^{k\ell} I(X,Z)$. Use the definition with ordered bases.

Ted Shifrin
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  • Hi Ted, how are you? The prop $I(Z,X)=(-1)^{k\ell} I(X,Z)$ is discussed on Page 115. But this remarked appeared on Page 112 asking to "compute the difference directly from the definition, assuming $X \pitchfork Z$." That's why I am attempting the more or less brutal method, trying to enumerate $8$ situations. So I am wondering if my approach is correct? Thank you Ted. – 1LiterTears Jul 21 '13 at 22:40
  • Oh no, I think I just need to note that the orientaion of $(a,b)$ is the negative of the orientation of $(b,a)$ if $a,b$ form a basis of the tangential space, hence complete my proof...? – 1LiterTears Jul 21 '13 at 22:46
  • Right. Because it takes $1$ switch to get from one to the other. – Ted Shifrin Jul 21 '13 at 22:48
  • Got it, thank you so much Ted! – 1LiterTears Jul 21 '13 at 22:49