# One form of poincare duality imply the other?

I saw in my differential geometry class that Poincare duality states:

$H_{DR}^k(X,\mathbb{R}) \times H_{DR}^{n - k}(X,\mathbb{R}) \rightarrow \mathbb{R}$ given by $([v],[r]) \mapsto \int_{X} n \wedge r$ is non-singular

Prof mentioned that If X is compact oriented manifold of dimension n, then this is really the same as saying:

$H_{DR}^i(X,\mathbb{R}) \cong H_{n - i}(X,\mathbb{R})$ where the right side of the equation we have singular homology. Can someone explain why this is the case ?

• It's quite a stretch to say that these two are "the same"... In fact it's flat out wrong, the first one just follows from the fact that $\mathbb{R}^n$ is contractible! Commented Dec 11, 2016 at 7:54
• If you assume the first can you prove the second one ?
– user329017
Commented Dec 11, 2016 at 8:00
• I mean I can prove the second one anyway, but it doesn't follow from the first one directly at all. Commented Dec 11, 2016 at 8:02
• Sorry @NajibIdrissi I was confusing something with something else. I have edited the question.
– user329017
Commented Dec 11, 2016 at 8:32

The first statement gives an isomorphism $H_{dR}^k(X,\mathbb{R})\to H_{dR}^{n-k}(X,\mathbb{R})^\vee$ where $^\vee$ denotes the dual vector space. By the de Rham theorem, $H^{n-k}_{dR}(X,\mathbb{R})\cong H^{n-k}(X,\mathbb{R})$ (where the right-hand side is singular cohomology), and by the universal coefficient theorem $H^{n-k}(X,\mathbb{R})\cong H_{n-k}(X,\mathbb{R})^\vee$. So putting these isomorphisms together, you get $H^k_{dR}(X,\mathbb{R})\cong (H_{n-k}(X,\mathbb{R})^\vee)^\vee\cong H_{n-k}(X,\mathbb{R})$.
Conversely, if you start with $H^k_{dR}(X,\mathbb{R})\cong H_{n-k}(X,\mathbb{R})$ you can reverse these steps and obtain an isomorphism $H^k_{dR}(X,\mathbb{R})\cong H_{dR}^{n-k}(X,\mathbb{R})^\vee$. However, this isn't quite as strong as your first statement, because it doesn't tell you exactly what the isomorphism in question is (namely, the isomorphism given by the perfect pairing you wrote).