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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    $\begingroup$ This makes very little difference... If you put $\tau$ on the right in the definition of $\delta$ then it's slightly more consistent, yes. $\endgroup$ – Najib Idrissi Jun 3 '19 at 5:56
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    $\begingroup$ 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. $\endgroup$ – Najib Idrissi Jun 3 '19 at 8:55
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    $\begingroup$ "Mistake" is rather excessive for what it is. $\endgroup$ – Najib Idrissi Jun 3 '19 at 15:49
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    $\begingroup$ I mean that yes, it is a mistake. And I mean that it is of no consequence. "Oversight", as I said, is more appropriate. $\endgroup$ – Najib Idrissi Jun 4 '19 at 7:27
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    $\begingroup$ @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. $\endgroup$ – Najib Idrissi Jun 5 '19 at 7:07

A definition is what one makes it to be. Sometimes there are choices and it is difficult to say whether one choice is more correct than another. There is virtue in consistency. In the discussion of Poincaré duality in DFAT, I put forms with no restrictions before forms with compact support. To be consistent with this, $\eta_S$ should be placed before (to the left of) $\omega \in H_c^k(M)$ and the closed Poincaré dual of a compact submanifold $S$ would be $\pm$ the compact Poincaré dual. However, one might decide to have the closed Poincaré dual equal to exactly the compact Poincaré dual. In this case, $\eta_S$ would be written after (to the right of) $\omega \in H_c^k(M)$. This was the choice I made. I don't think it is incorrect.

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    $\begingroup$ Welcome to MO, Professor Tu! $\endgroup$ – David Roberts Jun 19 '19 at 13:46
  • $\begingroup$ Thanks! Oh wait I think I get it now, Prof Tu. Do you mean $[\eta_S] := (-1)^{k} (-1)^{n-k} [\gamma_S] := [(-1)^{k} (-1)^{n-k} \gamma_S]$? Please see my edit to my question. $\endgroup$ – Selene Auckland Aug 5 '19 at 5:37
  • $\begingroup$ Thanks Prof Tu! $\endgroup$ – Selene Auckland Oct 22 '19 at 13:37
  • $\begingroup$ This is a response to Selena Auckland's edit of August 5 '19. When a k-form and an (n-k)-form are exchanged in a wedge product, the resulting sign is $(-1)^{k(n-k)}$, not $(-1)^k (-1)^{n-k}$. Otherwise, I think your edit is correct. $\endgroup$ – Loring Tu Feb 15 '20 at 22:28

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