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My book is Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott of which An Introduction to Manifolds by Loring W. Tu is a prequel.

The characterization of the closed Poincaré dual is given here (the "(5.13)") in Section 5.5. This has $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left.

Question: Why is it $\int_M \omega \wedge \eta_S$, where $\eta_S$ is on the right rather than left?

  • See below for why I think $\eta_S$ should be on the left rather than right.

  • Previously, in Section 5.3, we had this equivalent definition (the "Lemma") of a nondegenerate pairing between two finite-dimensional vector spaces and Poincaré duality (the "(5.4)").

  • Note: I believe the characterization for compact Poincaré dual for compact $S$ and $M$ of finite type is correct with $\eta_S'$ on the right.

  • Guess: Could have something to do with sign commutativity of Mayer-Vietoris, as described in Lemma 5.6.

  • Guess: Poincare dual as described is indeed with $\eta_S$ on the left, but there's also a unique cohomology class $[\gamma_S]$ that's on the right given by $[\gamma_S] = [-\eta_S]$.


How I got $\int_M \eta_S \wedge \omega$ instead of $\int_M \omega \wedge \eta_S$:

I use $()^{\vee}$, instead of $()^{*}$, to denote dual just like in Section 3.1 of the prequel.

  1. Let $\varphi$ be the "linear functional on $H^{k}_cM$" given here.

    • Such $\varphi: H^{k}_cM \to \mathbb R$ is given by $\varphi[\omega] = \int_S \iota^{*}\omega$ for $[\omega] \in H^k_cM$ and $\iota: S \to M$ inclusion.
  2. Let $\delta$ be the isomorphism of Poincaré duality (the "(5.4)").

    • Such $\delta: H^{n-k}M \to (H^{k}_cM)^{\vee}$ is given by $\delta([\tau]) = \delta_{[\tau]}$, for $[\tau] \in H^{n-k}M$ and $\delta_{[\tau]}$ given below.

    • $\delta_{[\tau]}([\omega]) = \int_M (\tau \wedge \omega)$, for $[\omega] \in H^k_cM$, under the well-definedness described in Section 24.4 of the prequel (which I think is the full details of the "Because the wedge product is an antiderivation, it descends to cohomology" here) and under the pairing given here, which I believe puts $\tau$ on the left rather than right.

  3. $[\eta_S]$ is the inverse image of $\varphi$ under $\delta$.

    • By choosing $[\tau] = [\eta_S]$, we get $\delta([\eta_S]) = \delta_{[\eta_S]} = \varphi$, that is, for all $[\omega] \in H^k_cM$,

$$\int_M (\eta_S \wedge \omega) = \int_S \iota^{*}\omega,$$

where $\eta_S$ is on the left rather than right.


Edit: After doing some thinking (it's easier to think when you know something is right/wrong as opposed to thinking about whether or not it's right/wrong, I believe), along with comments of Najib Idrissi and answer of Prof Tu, I think I've got it. Is this right?

We get a unique class $[\gamma_S]$ where for $\gamma_S \in [\gamma_S]$ (or any other element of $[\gamma_S]$), we have that for all $[\omega] \in H^k_cM$ $\omega \in [\omega]$ (or any other element of $[\omega]$),

$$\int_M (\gamma_S \wedge \omega) = \int_M ((-1)^{k} (-1)^{n-k}\omega \wedge \gamma_S) = \int_M (\omega \wedge (-1)^{k} (-1)^{n-k} \gamma_S) = \int_S \iota^{*}\omega$$ and then define $[\eta_S] := (-1)^{k} (-1)^{n-k} [\gamma_S] := [(-1)^{k} (-1)^{n-k} \gamma_S]$.

In this case, I think $[\eta_S] := - [\gamma_S] := [-\gamma_S]$ is a different definition from the one in the preceding paragraph unless $k(n-k)$ is an odd integer or something.

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    This makes very little difference... If you put $\tau$ on the right in the definition of $\delta$ then it's slightly more consistent, yes. – Najib Idrissi Jun 03 '19 at 05:56
  • @NajibIdrissi Do you really mean $\tau$ and not $\eta_S$? I actually think the one with $\tau$ is correct – Selene Auckland Jun 03 '19 at 07:02
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    I don't see why putting it on the left would be correct and on the right would be incorrect. In the end you're just changing a sign... The probable answer for "Why?" is certainly that the book is more than 300 pages long and this is an inconsequential oversight. – Najib Idrissi Jun 03 '19 at 08:55
  • @NajibIdrissi Wait so it's really a mistake? That's all I needed: either it's a mistake or I misunderstood somehow. – Selene Auckland Jun 03 '19 at 09:48
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    "Mistake" is rather excessive for what it is. – Najib Idrissi Jun 03 '19 at 15:49
  • @NajibIdrissi 1. What do you mean then? 2. My derivation for the closed Poincaré dual gives me $\eta_S$ on the left rather than right. Am I wrong? 3. The book's derivation gives $\eta_S$ on the right rather than left. Is that book wrong? 4. If you answered no to both (2) and (3), then do you mean $$\int_M (\eta_S \wedge \omega) = \int_S \iota^{}\omega$$ holds for each $\omega$ if and only if $$ \int_M (\omega \wedge \eta_S) = \int_S \iota^{}\omega$$ holds of reach $\omega$? I guess this is by substitution like $\int_0^1 x dx = \int_{-1}^0 -x dx$? – Selene Auckland Jun 03 '19 at 23:41
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    I mean that yes, it is a mistake. And I mean that it is of no consequence. "Oversight", as I said, is more appropriate. – Najib Idrissi Jun 04 '19 at 07:27
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    For the record, I think reading a value judgement into 'mistake' is (well) a mistake. If there is a typo on p. 363 of an otherwise-perfect Serre manuscript, say, then I would call it a mistake, but not mean any slight by it. I suspect the authors themselves would cheerfully call this an easily fixed mistake. – LSpice Jun 04 '19 at 14:50
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    @SeleneAuckland You realize that all it changes is a sign, right? In your other question, it's then $-\eta_S$ which is the closed Poincaré dual. – Najib Idrissi Jun 05 '19 at 07:07
  • @NajibIdrissi Thanks, but I think I'm confused by these mini-answer comments. I hope you could please clarify with a direct answer in an answer post. 1. So (5.13) is supposed to have $\eta_S$ on the left rather than the right? Please answer directly with a yes or no. 2. For (5.14), the compact Poincaré dual is not the closed Poincaré dual but rather the negative of the closed Poincaré dual? – Selene Auckland Jun 10 '19 at 05:36
  • Yes. 2. Yes.
  • – Najib Idrissi Jun 10 '19 at 06:51
  • @NajibIdrissi Edited question. Do we have $[\eta_S] := (-1)^{k} (-1)^{n-k} [\gamma_S] := [(-1)^{k} (-1)^{n-k} \gamma_S]$ rather than $[\eta_S] := [- \gamma_S]$? Or is $k(n-k)$ somehow odd? – Selene Auckland Aug 05 '19 at 06:07