Let $L$ be a finite dimensional vector space over some field $\mathbb{F}$ ($\mathrm{char}(\mathbb{F})=0$), $H\subset L$ a subspace of $L$ with $\mathrm{dim}_{\mathbb{F}}(H)=h$ and let $H^*$ be its dual space. Now let $\Phi\subset H^*$ be a root system$^1$ (aka $\Phi$ spans $H^*$) and define $$E_{\mathbb{Q}}:=\mathrm{span}(\Phi\mid \text{with coefficients in }\mathbb{Q}),$$ so we have $E_{\mathbb{Q}}\subset H^*$. We can now define $$\mathbb{E}:=\mathbb{R}\otimes_{\mathbb{Q}}E_{\mathbb{Q}}.$$ I'm having a hard time relating this space $\mathbb{E}$ to $H^*$. I think its basis should be $\{1\otimes\alpha\}_{\alpha\in E_{\mathbb{Q}}}$ and every element of this space is a linear combination of this basis with coefficients from $\mathbb{Q}$ (thats at least how I interpret $\otimes_{\mathbb{Q}}$).
Questions:
- In the lecture it was mentioned that $\mathbb{E}$ "is obtained by extending the base field from $\mathbb{Q}$ to $\mathbb{R}$". This sounds to me as if $\mathbb{E}\cong E_{\mathbb{R}}$, is this true?
- Assuming $\mathbb{F}=\mathbb{C}$, is $H^*\cong \mathbb{E} \oplus i \mathbb{E}$?
1) I initially left this information out, since I didn't think it was relevant for the question, but since it may be here is a bit more context: the question came up when I was studying section 8.5 of "Introduction to Lie Algebras and Representation Theory", J. E. Humphreys.
I'll try to answer all of your questions, step by step. First of all, contrary to what is claimed in the OP as of now, a root system is much more than just a set that spans $H^*$. This will have consequences in the answer to question 4 below.
To view the mathemical objects more clearly, you need to consider bases. In general, $\Phi$ will not be a ${\mathbb Q}$-basis of $E_{\mathbb Q}$, but we can extract a basis $\Psi \subseteq \Phi$ from it. We also need a $\mathbb Q$-basis $B$ of $\mathbb F$.
Then, we know that $\mathbb{E}(\mathbb{F})=\mathbb{F}\otimes_{\mathbb{Q}}E_{\mathbb{Q}}$ is a ${\mathbb{Q}}$-vector space with basis $\lbrace b\otimes_{\mathbb{Q}} \psi \ | \ b\in B,\psi \in \Psi \rbrace$.
Question 1: How is $\mathbb{E}(\mathbb{F})$ related to $H^*$ ?
Answer: Consider the map $f : B \times \Psi \to H^*$, defined by $f(b,\psi)=b\psi$. Then $f$ can be uniquely extended to a $\mathbb Q$-bilinear map $f^{\sim}: \mathbb{E}(\mathbb{F}) \to H^*$. Because both $B$ and $\Psi$ are bases in their respective spaces, $f^{\sim}$ is bijective. This allows us to identify $\mathbb{E}(\mathbb{F})$ with $f^{\sim}({\mathbb{E}}(\mathbb{F}))$ which is a $\mathbb Q$-subpace of $H^*$. Note that, since $f^{\sim}({\mathbb{E}}(\mathbb{F}))$ is stable by multiplication by an element of $B$ or $\mathbb Q$, it is also stable by multiplication by an element of $\mathbb F$, so it is a $\mathbb F$-subpace of $H^*$ as well.
Question 2: Is a ${\cal B}=\{1\otimes\alpha\}_{\alpha\in E_{\mathbb{Q}}}$ a basis of $\mathbb E(\mathbb{F})$ ?
Answer: No. It is not a basis because it is not linearly independent : for a non-zero $\alpha \in E_{\mathbb{Q}}$, if you put $v_1=1\otimes\alpha$, $v_2=1\otimes(-\alpha)$, then $v_1$ and $v_2$ are two distinct vectors of $\cal B$ but $v_1+v_2=0$.
On the other hand, it is a generating set over $\mathbb F$ (but not over $\mathbb Q$ in general).
Question 3: Does $\mathbb{E}\cong E_{\mathbb{R}}$ ?
Answer: (here ${\mathbb F}=\mathbb R$) Yes, because if you recall the $f^{\sim}$ from question 1, the image $f^{\sim}({\mathbb{E}})$, coincides with the $\mathbb R$-space of linear combinations with real coefficients of vectors in $\Psi$ . So $f^{\sim}({\mathbb{E}}) \cong E_{\mathbb{R}}$.
Question 4: Assuming $\mathbb{F}=\mathbb{C}$, is $H^*\cong \mathbb{E} \oplus i \mathbb{E}$?
Answer: (here ${\mathbb F}=\mathbb C$) Yes to the sum decomposition part, No to the direct sum part (although the sum is indeed direct when $\Phi$ is a root system). Since $\Phi$ generates $H^*$ over $\mathbb C$ and $\Psi$ generates $\Phi$ over $\mathbb Q$, it follows that $\Psi$ generates $H^*$ over $\mathbb C$ also. So the elements of $H^*$ are exactly the sum of elements of the form $v=(a+ib)\psi$, with $a,b\in{\mathbb R},\psi \in \Psi$. We can then write $v=a\psi+ib\psi$ ; the first summand is in $\mathbb E$ while the second is in $i\mathbb E$ ; this shows that $H^*= \mathbb{E} + i \mathbb{E}$.
However, there is no reason for the sum $\mathbb{E} + i \mathbb{E}$ to be direct : if, for example, you have a $\phi$ such that $\Phi$ contains both $\phi$ and $i\phi$, then $\phi \in \mathbb{E} \cap i \mathbb{E}$.
On the other hand, when $\Phi$ is a root system, its members live in a Euclidean space with real coordinates, so the sum is obviously direct.