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$\mathfrak{g}$ is a finite-dimensional complex semisimple Lie algebra and $\mathfrak{h}$ is one of its Cartan subalgebra.

$V$ is a (finite-dimensional?) complex vector space and $ρ:\mathfrak{g}\to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}$ on $V$.

A weight $λ$ is a linear functional on $\mathfrak{h}$ for which there exist a nonzero vector $v$ in $V$ so that for each element $H\in\mathfrak{h}$ there is $ρ(H)(v)=λ(H)v$.

A root is a nonzero weight when $ρ$ is the adjoint representation.

So weights and roots are both elements of $\mathfrak{h}^*$

Then according to the "Integral element" section of "Weight (representation theory)" article of Wikipedia,

Let $\mathfrak{h}_0^*$ be the real subspace of $\mathfrak{h}^*$ generated by the roots which is the dual space of the subspace $\mathfrak{h}_0$ of $\mathfrak{h}$.

So although $\mathfrak{h}_0^*$ is a real vector space, but its elements are complex linear functionals.

By choosing an inner product on $\mathfrak{g}$ (which seems to be the killing form), elements (including the roots and weights defined above) of $\mathfrak{h}_0^*$ can be identified to elements in $\mathfrak{h}_0$.

The identified roots and weights on $\mathfrak{h}_0$ have the properties of a root system.

The confusion is, as the space of root system, $\mathfrak{h}_0$ must be an Euclidean space i.e. real vector space, but its dual space $\mathfrak{h}_0^*$ consist of complex linear functionals...

So is there another implicit identification in $\mathfrak{h}_0^*$ not mentioned by the article?

jw_
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    You wrote:"Let $\mathfrak{h}0^$ be the real subspace of $\mathfrak{h}^$". So it is a real vector space. You can even consider it over $\Bbb Q$, so then $\dim{\Bbb Q} \mathfrak{h}{\Bbb Q}^*=\dim{\Bbb C} \mathfrak{h}^$. Each root can be uniquely represented by a rational linear combination of the basis elements. We usually consider the Euclidean space $E=\mathfrak{h}_{\Bbb R}^$ generated by the roots as a real vector space. Then the Killing form yields a positive definite scalar product $(\lambda,\mu)$. – Dietrich Burde Jan 10 '23 at 12:39
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    What Dietrich said. Also, related: https://math.stackexchange.com/q/3312731/96384 – Torsten Schoeneberg Jan 10 '23 at 17:27

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