Poincaré duality and Hodge duality

Question

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 $a\in \bigwedge (\mathbb R^n)^*$.

EDIT:

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.

Answer 1

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.

Answer 2

$ \newcommand\R{\mathbb R} \newcommand\Ext\bigwedge \newcommand\form[1]{\langle#1\rangle} \newcommand\rev\widetilde \newcommand\hodge\star $

The punchline is that $$ \hodge X = D_\#(Q(X))\qquad \forall X\in\Ext\R^n, $$ where $\hodge$ is the Hodge star, $D_\# : \Ext(\R^n)^* \to \Ext\R^n$ is one of Greub's Poincaré isomorphisms, and $Q : \Ext\R^n \to \Ext(\R^n)^*$ is the isomorphism induced by the non-degenerate bilinear form used to define $\hodge$.

---

Choosing a non-degenerate bilinear form on $\R^n$ amounts to a choice of isomorphism $Q : \R^n \to (\R^n)^*$, and upon making this choice $\Ext\R^n$ inherits the Clifford product of the associtated Clifford algebra. By the universal property of Clifford algebras $Q$ extends to a Clifford algebra isomorphism $\Ext\R^n \to \Ext(\R^n)^*$ where we give $\Ext(\R^n)^*$ the dual Clifford product defined by $Q^{-1}$. Writing Clifford multiplication as juxtaposition, for $X, Y \in \Ext\R^n$ we get $$ \form{Q(X), Y} = \form{\rev XY}_0, $$ where $\rev X$ is the reversal (i.e. main anti-automorphism) and $\form{\cdot}_0$ is scalar projection.

One of the Poincaré isomorphisms defined by Greub is $$ D_\# : \Ext(\R^n)^* \to \Ext\R^n,\quad D_\#(\xi) = \iota_\xi(I) $$ where $\iota$ is the interior product and for some choice of $I \in \Ext^n\R^n$, which with our Clifford product we will choose to be a unit $I^2 = \pm1$ to line up with the Hodge star later. For $A \in \Ext^{n-k}\R^n$ and $B \in \Ext^{k}\R^n$, we see that the manifestation of $D_\#$ in the Clifford algebra is $$\begin{aligned} \form{\rev A(D_\#\circ Q(B))}_0 &= \form{Q(A), D_\#(Q(B))} \\ &= \form{Q(A), \iota_{Q(B)}(I)} \\ &= \form{Q(B)\wedge Q(A), I} \\ &= \form{Q(B\wedge A), I} \\ &= \form{\rev{B\wedge A}\:I}_0 \\ &= \form{\rev A\wedge\rev B\:I}_0 \\ &= \form{\rev A(\rev BI)}_0 \end{aligned}$$$$ \implies D_\#\circ Q(B) = \rev BI, $$ with the last line following since $A$ is arbitrary and $\form{\rev{(\cdot)}\,{\cdot}}_0$ is a non-degenerate bilinear form. The step $Q(B)\wedge Q(A) = Q(B\wedge A)$ is valid since $\wedge$ can be realized with the Clifford product, e.g. $A\wedge B = \form{AB}_n$ with $\form{\cdot}_n$ the projection onto $\Ext^n\R^n$. Since we proved the above for simple $k$-vectors, it holds for all multivectors: $$ D_\#\circ Q(X) = \rev XI\qquad\forall X\in\Ext\R^n. $$

Now consider the Hodge star, which following this is defined by $$ A\wedge(\hodge B) = \form{\rev AB}_0I $$ for $A, B \in \Ext^k\R^n$. Converting the RHS to a $\wedge$ and multiplying by $I$ on the left, we immediately get $$ I\:A\wedge(\hodge B) = I\:A\wedge(\rev BI), $$ where we've also move the reversal onto $B$ since both $A$ and $B$ are $k$-vectors. Then $$ \form{(IA)(\hodge B)}_0 = \form{(IA)(\rev BI)}_0, $$ and it follows by non-degeneracy of the form that $$ \hodge B = \rev BI $$ and this result extends to all multivectors. We have finally proven that $$ \hodge X = D_\#(Q(X))\qquad \forall X\in\Ext\R^n. $$