The Poincaré duality is defined in Greub's Multilinear algebra (1967) in Chapter 6, §2 as a isomorphism between $\bigwedge V$ and $\bigwedge V^*$, where $V$ is a finite-dimensional vector space, $V^*$ is its dual, and $\bigwedge V$ is the exterior algebra of $V$. More precisely, the isomorphism maps $\bigwedge^k V$ to $\bigwedge^{n-k} V^*$, $n=\dim V$. Greub's definition has nothing to do with cohomology groups of manifolds and is based entirely on multilinear algebra. I have tried reading Chapter 6 and quickly realised I would need to study the entire book to understand the definition. I am only interested in the cases $V=\mathbb R^n$ where $n\le5$, and I only need to see what the isomorphism looks like on the standard basis of $\bigwedge \mathbb R^n$ and $\bigwedge (\mathbb R^n)^*$.

For example, let $e_1,e_2,e_3$ denote the standard basis of $\mathbb R^3$ and $e_1^*,e_2^*,e_3^*$ the standard basis of $({\mathbb R}^3)^*$. Then the basis of $\bigwedge\mathbb R^3$ consists of 1, $e_1,e_2,e_3$, $e_1\wedge e_2$, $e_2\wedge e_3$, $e_3\wedge e_1$, $e_1\wedge e_2\wedge e_3$ and the basis of $\bigwedge(\mathbb R^3)^*$ consists of 1, $e_1^*,e_2^*,e_3^*$, $e_1^*\wedge e_2^*$, $e_2^*\wedge e_3^*$, $e_3^*\wedge e_1^*$, $e_1^*\wedge e_2^*\wedge e_3^*$. The isomorphism maps, say, $e_1^*$ to $e_2\wedge e_3$ or perhaps $-e_2\wedge e_3$. I am not sure which sign is the right one. That is, I already know what the isomorphism looks like but only up to a sign. So, my question is what the correct signs are in the mapping Greub uses.

My second question is how Poincaré duality as defined by Greub is related to Hodge duality. The latter depends on a bilinar form on $(\mathbb R^n)^*$ but I am only interested in the form defined by $\langle e_i^*,e_i^*\rangle=1$ and $\langle e_i^*,e_j^*\rangle=0$ if $i\ne j$. The Hodge duality maps $\bigwedge^k(\mathbb R^n)^*$ to $\bigwedge^{n-k}(\mathbb R^n)^*$, so it is restricted to the exterior algebra of $(\mathbb R^n)^*$, but its effect is somewhat similar to that of the Poincare duality. For instance, the Hodge duality maps $e_1^*$ to $e_2^*\wedge e_3^*$. If I define a map $\mathcal{R}:\bigwedge (\mathbb R^n)^*\to\bigwedge \mathbb R^n$ in the obvious way by removing $*$ from the relevant basis vectors, e.g. $\mathcal{R}(e_1^*\wedge e_2^*)=e_1\wedge e_2$, would the Poincaré duality $\mathcal{P}:\bigwedge (\mathbb R^n)^*\to\bigwedge \mathbb R^n$ and the Hodge duality $\mathcal{H}$ be related by $\mathcal{P}(a)=\mathcal{R}(\mathcal{H}(a))$ where $a\in \bigwedge (\mathbb R^n)^*$.


Despite what Bruno Joyal says, I think $\mathcal{P}(a)=\mathcal{R}(\mathcal{H}(a))$ is valid without any sign adjustment, provided that $\langle,\rangle$ is defined as above. I define the Hodge dual (or Hodge star *, $\mathcal{H}(a)=*a$) by

$a\wedge(*b)=\langle a,b\rangle \omega$,

where $\omega=e_1^*\wedge\dots\wedge e_n^*$ is the standard element of $\bigwedge^n(\mathbb R^n)^*$, and $\langle a,b\rangle=\det(\langle a_i,b_i\rangle)$ for $a=a_1^*\wedge\dots\wedge a_k^*$ and $b=b_1^*\wedge\dots\wedge b_k^*$, which can be replaced by 1 in calculations with the basis vectors thanks to my choice of the inner product.

I don't understand the definition of Poincaré duality but Greub does give the formula for computing the duals,

$\mathcal{P}(e_{\nu_1}^*\wedge\dots\wedge e_{\nu_p}^*)=(-1)^{\sum_{i=1}^p(\nu_i-i)}e_{\nu_{p+1}}\wedge\dots\wedge e_{\nu_n}$,

where $\nu_1<\nu_2<\dots < \nu_n$, and $(\nu_{p+1},\dotsc,\nu_n)$ is complementary to $(\nu_1,\dots,\nu_p)$. As far as I can see, the factor $(-1)^{\sum_{i=1}^p(\nu_i-i)}$ is the sign of the permutation that takes $(\nu_1,\dots,\nu_n)$ to $(1,\dots,n)$, but $a$ and $*a$ are related in the same way for the basis elements of $\bigwedge(\mathbb R^n)^*$. I just need to use the map $\mathcal{R}$ to go to the primary space. Hence, $\mathcal{P}(a)=\mathcal{R}(\mathcal{H}(a))$. I confirmed that by calculations for $n=2,3,4$. The fact that $*a$ is not an involution doesn't really matter.

  • $\begingroup$ It would be helpful if someone could confirm my conclusion that $\mathcal{P}(a)=\mathcal{R}(\mathcal{H}(a))$ for the standard Euclidean bilinear form $\langle,\rangle$. $\endgroup$ – Andrey Sokolov Nov 8 '13 at 5:39

For Poincaré duality, it is normal that you are confused by the sign. The isomorphism $\Lambda^k V \to \Lambda^{n-k} V^*$ depends on the choice of an orientation form for $V$. Namely, we know that $\Lambda^nV \cong \mathbf R$, but generally there is no preferred isomorphism. The choice of such an isomorphism is equivalent to the choice of a nonzero element of $\Lambda^n V$, which we call an orientation form. Now, fix such a form. Wedge product determines a nondegenerate bilinear map

$$\Lambda^kV \times \Lambda^{n-k}V \to \Lambda^n V \simeq \mathbf R,$$ and this map is the one which gives Poincaré duality: it identifies $\Lambda^kV$ with $(\Lambda^{n-k}V)^* = \Lambda^{n-k}V^*$.

So, if you choose $e_1 \wedge e_2 \wedge e_3$ as your orientation form (which is customary), then under Poincaré duality, $e_1 \mapsto e_2 \wedge e_3$, $e_2\mapsto -e_1 \wedge e_3$ and $e_3 \mapsto e_1 \wedge e_2$.

As for Hodge duality, it is a bit more subtle. As you say, its construction depends on an inner product. If you want to write down Hodge duality for $\mathbf R^n$ with its usual inner product, you should be able to relate it to Poincaré duality up to a sign. Indeed, Hodge duality is not an involution; when $k$ and $n-k$ are both odd, its square is equal to $-1$ times the identity. (Of course, this never happens if $n$ is odd.) Poincaré duality, however, is always an involution.

  • $\begingroup$ Thanks, this helps. If possible, could you give a bit more detail about how Poincaré duality is constructed. I understand what you said but only superficially I'm afraid. Greub does define a bilinear map on the tensor product of exterior algebras, but I couldn't quite get past all the technical details. $\endgroup$ – Andrey Sokolov Nov 6 '13 at 6:12
  • $\begingroup$ @AndreySokolov, you are welcome. I will do that with pleasure, but it'll have to be tomorrow, as I must get some sleep! $\endgroup$ – Bruno Joyal Nov 6 '13 at 6:14
  • $\begingroup$ @AndreySokolov Ok. I can provide more details if you'd like, but you should indicate where exactly you would like me to supply details. $\endgroup$ – Bruno Joyal Nov 7 '13 at 3:18
  • $\begingroup$ Here would be fine. Did you have anything else in mind? $\endgroup$ – Andrey Sokolov Nov 7 '13 at 7:10
  • $\begingroup$ Dear @AndreySokolov: I meant to say that you should ask a more precise question because I'm not sure what to tell you. $\endgroup$ – Bruno Joyal Nov 8 '13 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.