What does it mean that inner product is bilinear and non-singular

[rayman123]

Homework Statement

Hello everyone! I am reading abour Poincares duality between 2 cohomology groups and here comes an inner product defined as

$\left\langle \omega, \eta\right\rangle \equiv \int_{M}\omega \wedge \eta$

and then the author of my book says ''The product is bilinear and non-singular'' that is if $$\omega\neq 0$$ or [itex] \eta \neq 0[/tex] , [itex]\left\langle \omega, \eta\right\rangle [/tex] cannot vanish identically...
can someone please explain me this concept, what does it mean bilinear and non-singular??

Thank you!

 

[HallsofIvy]

A function, f(u, v), is "bilinear" if and only if f(au+ bw,v)= af(u, v)+ bf(w,v) and f(u, av+ bw)= af(u,v)+ bf(u,w).

Non-singular simply means that f(u,v) is not 0 for all u and v.

 

[rayman123]

A function, f(u, v), is "bilinear" if and only if f(au+ bw,v)= af(u, v)+ bf(w,v) and f(u, av+ bw)= af(u,v)+ bf(u,w).

okej so f is a function of 2 variables, and are a and b some constants? I am not sure If I understand this equation...
Oh now I think I am getting it

 

[rayman123]

I have another question about the very same wedge product

in my book it is written
[tex][\omega \wedge \eta][/itex] is independent of the choice of the representatives of $$[\omega]$$ and [itex] [\eta][/tex]. For example if we take [itex] \omega^{'}=\omega+d\psi[/tex] instead of [itex] \omega[/tex] we have

[itex] [\omega^{'}\wedge \eta]\equiv [(\omega+d\psi)\wedge \eta]=[\omega\wedge \eta+d(\psi\wedge \eta)]=[\omega\wedge \eta][/tex]


then the term $$d(\psi\wedge \eta)=0$$ but why
I checked the definition for the wedge product $$d(\omega\wedge \eta)=(d\omega)\wedge \eta+ (-)^{q}\omega\wedge (d\eta)$$ where [itex] \omega\in \Omega^{q}(M)[/tex] and [itex] d\omega\in \Omega^{q+1}(M)[/tex]


But I cannot see it giving me 0...Could you show it to me how did they get $$d(\psi\wedge \eta)=0$$