0
$\begingroup$

Consider $X= \mathbb{R}^2-\{(0,0),(1,0),(2,0),(3,0)\}.$ Let $V$ be the vector space of irrotational vector fields over $X.$ Let $W$ be the vector space of conservative vector fields over $X.$ What is the dimension of $V/W?$

So far I have deduced through basic definitions that Irrotational $\implies$ conservative (provided that the domain is simply connected.) Since the domain is not simply connected we may not say $V$ is necessarily conservative, which means the curl is not necessarily $0.$ Are these assumptions correct? If so, then I have also referred to another question on this website with which de Rham cohomology was used to answer the dimension of the quotient vector space, which was $1$, because the had one "hole" because the domain was $\mathbb{R}-\{(0,0)\}$. In this case, there are four holes so I imagined the dimension may be $4$, but I don't believe the answer is as straightforward as that and it wouldn't make sense either because we are in $\mathbb{R}^{2}$. I was hoping someone could help me understand how to solve this problem using "calculus and linear algebra basics," as OP suggests in a comment thread of their post. Alternatively, I would require some background information and a thorough explanation if de Rham cohomology is used because I have not yet reached that level in my studies.

$\endgroup$
1
$\begingroup$

An irrotational vector field ${\bf K}$ defined on $X$ has ${\rm curl}({\bf K})\equiv 0$. But we all know that there are irrotational fields in $\dot{\mathbb R}$ that are not conservative, the simplest being the field $${\bf A}(x,y)=\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},\>{x\over x^2+y^2}\right)\ ,$$ having integral $\int_\gamma{\bf A}\cdot d{\bf z}=2\pi$ for closed curves circling the origin once counterclockwise. Using this ${\bf A}$ define the translated fields $${\bf A}_k(x,y):={\bf A}(x-k,y)\qquad(0\leq k\leq 3)\ .$$ Given any field ${\bf K}\in V$ define the numbers $$c_k:={1\over 2\pi}\int_{\gamma_k}{\bf K}\cdot d{\bf z}\qquad(0\leq k\leq 3)\ ,$$ whereby $\gamma_k$ is a small circle around the point $(k,0)$. Then the field $${\bf L}:={\bf K}-\sum_{k=0}^3 c_k{\bf A}_k$$ has integral $0$ around all $\gamma_k$, hence is conservative. This allows to conclude that ${\rm dim}(V/W)=4$.

$\endgroup$
  • $\begingroup$ Thank you for your answer, it was very clear and concise. $\endgroup$ – JohnColtraneisJC Apr 26 '18 at 20:20

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.