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Let $V=\mathbb{C}\backslash \left\{-1\right\}$ and define the special addition $\oplus$ on $V$, for $\forall a,b\in V$ $$a\oplus b=a+b+ab.$$

Then we can check that $V$ satisfies all additive conditions in linear space with the special addition $\oplus$.

For example, the zero element is $0$, the additive inverse element of $a$ is $\dfrac{-a}{a+1}$.

My question is how can I define the scalar multiplication on $V$ so that $V$ can be seen as a $\mathbb{C}-$ linear space?

Any help and references are greatly appreciated.

Thanks!

fusheng
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    The operation that you define does not satisfy distributivity under scalar multiplication. Indeed, $r(a \oplus b) = ra + rb + rab)$ but $ra \oplus rb = ra + rb + r^2 b$. As $r \neq r^2$ in general for real numbers $r$, the binary operation does not satisfy the axioms for a real vector space. In particular, it cannot be a complex vector space. – kesa Mar 14 '24 at 08:47
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    More trivially, the space is not closed under scalar multiplication, as $(-1). 1 = -1$ – kesa Mar 14 '24 at 08:50
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    More generally, one can easily see that no real vector space structure can exist on $\mathbb{C} \setminus {x}$. Indeed, any such vector space would be finite dimensional (by cardinality argument), hence linearly isomorphic to some $\mathbb{R}^n$. However, $\mathbb{R}^n$ is contractible (or homotopy equivalent to a point) but $\mathbb{C} \setminus {x}$ is not (neither). As these properties would be inherited through a linear isomorphism, we get a contradiction. – kesa Mar 14 '24 at 08:54
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    Compare Ring isomorphism via $,a\oplus b = a + b + 1,\ \color{#c00}{a\odot b = a\cdot b + a + b},$ (transport of structure) $\ \ $ – Bill Dubuque Mar 14 '24 at 17:39
  • @kesa Thanks! I know the usual scalar multiplication does not satisfy the conditions. I want to find the special scalar multiplication so that $V$ can be seen as a real vector space. – fusheng Mar 15 '24 at 11:15
  • Still, then the general statement about $\mathbb{R}$ vector space structure on $\mathbb{C} \setminus x$ applies. It is independent of how the $\mathbb{R}$-action on $\mathbb{C} \setminus x$ is defined. Of course, one way of putting a vector space structure on $\mathbb{C} \setminus x$ is possible via the choice of a bijection between $\mathbb{C} \setminus x$ and $\mathbb{C}$, however, none such can be continuous in the usual topology and the resulting $\mathbb{R}$ structure will be as ugly as the bijection. I don't know, if more can be said about this. – kesa Mar 16 '24 at 12:48

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