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I am confused about a statement about Intersection forms (for $4$-manifolds):

[...] Define the intersection form modulo 2 $${\displaystyle \cap _{M,2}:H_{2}(M;\mathbb {Z} /2\mathbb {Z} )\times H_{2}(M;\mathbb {Z} /2\mathbb {Z} )\to \mathbb {Z} /2\mathbb {Z} }$$

by the formula

$${\displaystyle a\cap _{M,2}b=|A\cap B|\mod 2}$$

This is well-defined because the intersection of a cycle and a boundary consists of an even number of points (why?) [...].

I not understand why an intersection of a cycle and a boundary consists of an even number of points. why is it true?

For example if we take two $3$-cells aka tetrahedrons $A,B$, which share exactly one common face. their images under the boundary operator are alternating sums $\sum_{i=0}^3 (-1)^{i}dA_i$ and $\sum_{i=0}^3 (-1)^{i}dB_i$, where by $dA_i,dB_j$ we denote the $i$-th and $j$-th faces (=$2$-cells) of $A$ and $B$.

by construction $\sum_{i=0}^3 (-1)^{i}dA_i$ and $\sum_{i=0}^3 (-1)^{i}dB_i$ are boundaries and therefore also $2$-cycles. but as $A$ and $B$ share exactly one common face, there exist a $dA_i$ and $dB_j$ with $dA_i=dB_j$ and the other faces are disjunct. doesn't this contradict to the statement above that the intersection of a cycle and a boundary consists of an even number of points?

  • When one talks about and counts intersections, one assumes that the creatures have been made transverse in the first place. Because a compact $1$-manifold with boundary has an even number of boundary components, when you talk about the intersection of a $2$-cycle with a $2$-boundary, you apply my remark to the $2$-cycle and the $3$-chain to get a union of closed intervals. At any rate, in your example, to talk about the intersection you must first homotop one of the identical faces to make it transverse to the other. – Ted Shifrin Nov 23 '19 at 19:30
  • @TedShifrin:when you talk about that two chains are transverse to each other, you mean that their intersection is a discrete (thus finite by compactness) set, or in other words, that their intersection not contain an $n$-simplex with $n >0$? –  Nov 23 '19 at 20:13
  • It depends on dimensions, of course. But in your PL setting, the $k$-faces should intersect the $\ell$-faces in dimension $k+\ell-\dim M$. Faces of complementary dimension should intersect in a finite set of points. – Ted Shifrin Nov 23 '19 at 20:41
  • @TedShifrin:I think I got it: to intersect a $2$-cycle with a $2$-boundary transversally coinsides with taking boundary of the intersection of the $2$-cycle from above with the $3$-chain, which maps to the considered $2$-boundary. and this intersection is by transitivity nothing, but the finite union of lines (=$1$-manifolds), thus their boundary consists of even number of points. is this exactly the argument you explained above? –  Nov 23 '19 at 21:01
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    Yes, precisely, but I was confused by the comma after "nothing." :) – Ted Shifrin Nov 23 '19 at 21:07
  • @TedShifrin: Is there any equivalent/"relative" of the intersection form for odd-dimensional manifolds? – MSIS Sep 29 '21 at 20:45
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    @MSIS The intersection form is on a fixed (co)homology group (the middle dimension). How are you proposing to define an intersection number of two classes whose dimensions do not add up to the ambient dimension? There are linking numbers in greater generality, but I can't imagine what you're thinking of. – Ted Shifrin Sep 29 '21 at 20:54
  • @TedShifrin: Thank you. May I ask you a general (Geometric) Topology question? I asked it here but have not gotten reply in a while : https://math.stackexchange.com/questions/3655758/trivial-homology-and-embeddings – MSIS Sep 29 '21 at 21:12

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