# The Dimension of a Quotient Vector Space

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.

## 1 Answer

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 ${\rm dim}(V/W)=4$.

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