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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:

  1. 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?
  2. 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.

Sito
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  • Are you assuming that the field $\mathbb F$ has characteristic $0$ (and therefore includes $\mathbb Q$)? Otherwise, I don't understand the definition of $E_{\mathbb Q}$. – Andreas Blass Feb 05 '20 at 18:44
  • @AndreasBlass Yes, I'm sorry. That's an assumption I forgot to mention... It was stated at the beginning of the lecture and never mentioned again afterwards.. I'll correct it directly – Sito Feb 05 '20 at 18:46
  • Minor criticism : since you have an $H$ and an $L$, the name I would have chosen for $\dim(H)$ would have been $h$ rather than $l$ :-) – Ewan Delanoy Feb 09 '20 at 16:05
  • @EwanDelanoy Thanks for pointing this out, changed! – Sito Feb 09 '20 at 18:55
  • What is the point of $L$ in this setting? – Captain Lama Feb 09 '20 at 19:06
  • Also, is $\Phi$ just a generating set, and not a basis? Because this changes a lot of things (if $\Phi=H^$ then $E_\mathbb{Q}=H^$ for instance). – Captain Lama Feb 09 '20 at 19:13
  • @CaptainLama The context would be "Introduction to Lie Algebras and Representation Theory", J. E. Humphreys, section 8 (this question came about when reading section 8.5). In this context $L$ is a Lie algebra, Prop. 8.3 clearly states that $\Phi$ spans $H^*$ and in section 10.1 the author shows that a basis can be constructed from this set, so in itself it is not a basis (since for every $\alpha\in\Phi$ also $-\alpha\in\Phi$, for example). Did I omit some necessary information? If so, could you maybe point out what exactly, so that I can edit the question? – Sito Feb 09 '20 at 19:24
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    I think you should just explain in your question that you are talking about root systems. It is a very well-known construction, and it is quite specific ($\Phi$ is indeed not a basis, but it is very far from being arbitrary). – Captain Lama Feb 09 '20 at 20:08
  • @CaptainLama I edited the question, if there is still something missing, please let me know. – Sito Feb 09 '20 at 20:24
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    While the accepted answer answers the question as is, allow me to advertise my answers to https://math.stackexchange.com/q/2746923/96384 and https://math.stackexchange.com/q/3312731/96384 which talk about why the root system is visualised in a Euclidean (i.e. real) vector space, and how little that vector space has to do with the "original" $\mathbb F$-vector spaces like $H$ or $H^*$. – Torsten Schoeneberg Feb 24 '20 at 08:39

1 Answers1

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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.

Ewan Delanoy
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  • In your answer to question 1 you seem to assume that $H^*$ is an $\mathbb R$-vector space. That's not assumed in the question. (And it is true only if $\mathbb F = \mathbb R$ or $\mathbb C$, and although these are the most common fields over which to study Lie algebras, they are of course not the only ones.) – Torsten Schoeneberg Feb 24 '20 at 08:46
  • @TorstenSchoeneberg No, I only use the fact $H^*$ is an $\mathbb F$-vector space where $\mathbb F$ has characteristic zero (as explicitly specified in the question). – Ewan Delanoy Feb 24 '20 at 10:45
  • Then I don't understand (if, say, $\mathbb F=\mathbb Q$) what "$\mathbb R$-sub[s]pace of $H^\ast$" is supposed to mean, and how for $b \in \mathbb R \setminus \mathbb Q$ the element $b\psi \in H^\ast$ (in your definition of $f$) is defined. – Torsten Schoeneberg Feb 24 '20 at 16:56
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    @TorstenSchoeneberg $\mathbb F$ is arbitrary in what I call questions 1 and 2, and becomes specialized to ${\mathbb F}={\mathbb R}$ in question 3, and specialized to ${\mathbb F}={\mathbb C}$ in question 4. I've edited my answer to make that clearer, thx for the correction – Ewan Delanoy Feb 24 '20 at 17:26
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    Thanks for correcting. Maybe it's worth pointing out that the object you now call $\mathbb E(\mathbb F)$ was never mentioned in the OP (the one they call $\mathbb E$ is only $\mathbb E(\mathbb R)$), and in case $\mathbb R \not\subseteq \mathbb F$ is distinctly different from $H^\ast$, starting with them being vector spaces over different fields. – Torsten Schoeneberg Feb 24 '20 at 18:58
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    @TorstenSchoeneberg Regarding ${\mathbb E}({\mathbb F})$, it's true that it's not considered directly in the OP, but I think it allows for a more thorough answer. Regarding the vector spaces over different fields, I think you're wrong : each of $H$, $H^{*}$ and ${\mathbb E}({\mathbb F})$ are vector spaces over $\mathbb F$, whatever $\mathbb F$ may be. – Ewan Delanoy Feb 25 '20 at 08:43
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    You're right, I misphrased in my last comment. I wanted to say that unless $\mathbb F$ contains the reals, the $\mathbb E$ of the OP is distinctly different from $H$ etc. – Torsten Schoeneberg Feb 25 '20 at 15:59
  • @TorstenSchoeneberg Agreed. – Ewan Delanoy Feb 25 '20 at 17:40