Understanding the Wedge Product on a 3-dim Manifold

[Silviu]

Hello! The cohomology ring on an M-dim manifold is defined as $H^*(M)=\oplus_{r=1}^mH^r(M)$ and the product on $H^*$ is provided by the wedge product between cohomology classes i.e. $[a]$ $\wedge$ $[c]$ $= [a \wedge c]$, where $[a]\in H^r(M)$, $[c]\in H^p(M)$ and $[a \wedge c]\in H^{r+p}(M)$. Can someone write down for me how does the wedge product act between 2 elements of $H^*$? Let's say the dim of M is 3 and we want to take the wedge product of 2 elements. The elements would be $(a^1,a^2,a^3)$ and $(c^1,c^2,c^2)$, with $a^i \in H^i(M)$ and same for c. Is this right? Now the wedge product would be $(a^i \wedge c^i)$? And how you define it when the upper index is greater than the dim of M? Thank you!

 

[lavinia]

The De Rham cohomology group $H^{i}(M)$ is the quotient group of closed $i$-forms modulo exact $i$-forms.

Two closed ${i}$-forms $α_1$ and $α_2$ represent the same cohomology class if $α_1 = α_2 + dω$ where $ω$ is an $i-1$ form and $dω$ is its exterior derivative.

Since these cohomology classes are elements of a quotient group one cannot take their wedge product. The best one can do is is to take the wedge product of differential forms that represent the cohomology classes. If $a$ and $b$ are elements of $H^{i}(M)$ and $H^{j}(M)$ and if $α$ and $β$ are differential forms representing $a$ and $b$ then one can take their wedge product $α∧β$ to get an $i+j$ form.

Since $α$ and $β$ are closed, $α∧β$ is also closed so it represents some cohomology class in $H^{i+j}(M)$. One defines the "cup product" on the cohomology groups as $a∪b= [α∧β]$ where $[α∧β]$ is the cohomology class of $α∧β$.

One must show that this product is well defined. If $α+dω$ is another differential form representing the cohomology class $a$ one must show that $[α∧β]= [(α+dω)∧β]$.

- If $i+j$ is greater than the dimension of the manifold $M$ then $α∧β$ is zero. This is because the tangent space to $M$ is an n-dimensional vector space at each point of $M$.

 

[Silviu]

The De Rham cohomology group $H^{i}(M)$ is the quotient group of closed $i$-forms modulo exact $i$-forms.

Two closed ${i}$-forms $α_1$ and $α_2$ represent the same cohomology class if $α_1 = α_2 + dω$ where $ω$ is an $i-1$ form and $dω$ is its exterior derivative.

Since these cohomology classes are elements of a quotient group one cannot take their wedge product. The best one can do is is to take the wedge product of differential forms that represent the cohomology classes. If $a$ and $b$ are elements of $H^{i}(M)$ and $H^{j}(M)$ and if $α$ and $β$ are differential forms representing $a$ and $b$ then one can take their wedge product $α∧β$ to get an $i+j$ form.

Since $α$ and $β$ are closed, $α∧β$ is also closed so it represents some cohomology class in $H^{i+j}(M)$. One defines the "cup product" on the cohomology groups as $a∪b= [α∧β]$ where $[α∧β]$ is the cohomology class of $α∧β$.

One must show that this product is well defined. If $α+dω$ is another differential form representing the cohomology class $a$ one must show that $[α∧β]= [(α+dω)∧β]$.

- If $i+j$ is greater than the dimension of the manifold $M$ then $α∧β$ is zero. This is because the tangent space to $M$ is an n-dimensional vector space at each point of $M$.

Thank you for your reply. However I understood this part. What I am confuse about is the actual multiplication of 2 elements of the ring. what is the results of $a \wedge c$, for a and c $\in H^*(M)$

 

[lavinia]

Thank you for your reply. However I understood this part. What I am confuse about is the actual multiplication of 2 elements of the ring. what is the results of $a \wedge c$, for a and c $\in H^*(M)$

I explained the multiplication.

 

[Silviu]

I explained the multiplication.

You explained the multiplication of 2 elements of 2 cohomology groups. I would like to know the multiplication of 2 elements of a cohomology ring i.e. the elements of a cohomology ring are not just cohomology classes but direct sum of these and I am not sure how does that looks explicitly.

 

[lavinia]

You explained the multiplication of 2 elements of 2 cohomology groups. I would like to know the multiplication of 2 elements of a cohomology ring i.e. the elements of a cohomology ring are not just cohomology classes but direct sum of these and I am not sure how does that looks explicitly.

Whenever one has a ring the product $(Σ_{i}a_{i})⋅(Σ_{j}b_{j})$ equals $Σ_{ij}a_{i}⋅b_{j}$. This rule tells you how to multiply arbitrary elements of the De Rham cohomology ring.

 

[Silviu]

Whenever one has a ring the product $(Σ_{i}a_{i})⋅(Σ_{j}b_{j}) = Σ_{ij}a_{i}⋅b_{j}$. This rule tells you how to multiply arbitrary elements of the De Rham cohomology group.

So assuming we have a 2D manifold an element of a cohomology ring would be $a_1+b_1dx+c_1dx\wedge dy$ and multiplying 2 elements would give $(a_1+b_1dx+c_1dx\wedge dy)\wedge (a_2+b_2dx+c_2dx\wedge dy) = a_1a_2+a_1b_2dx+a_1c_2dx\wedge dy + b_1a_2dx+c_1a_2dx\wedge dy$, while all the other terms get to 0? Is this correct?

 

[lavinia]

So assuming we have a 2D manifold an element of a cohomology ring would be $a_1+b_1dx+c_1dx\wedge dy$ and multiplying 2 elements would give $(a_1+b_1dx+c_1dx\wedge dy)\wedge (a_2+b_2dx+c_2dx\wedge dy) = a_1a_2+a_1b_2dx+a_1c_2dx\wedge dy + b_1a_2dx+c_1a_2dx\wedge dy$, while all the other terms get to 0? Is this correct?

Unless you made an algebraic mistake this looks right.