I don´t understand root systems

Question

I don´t understand root systems. The Wikipedia (and my university lectures) say it is some configuration of vectors with certan properties. The root vectors should span the whole space, which I imagine as generating the space (making linear combination of the root vectors).

So does that mean that these root vectors have linear span or are something like a basis?

I fail to grasp even the most simple example on Wikipedia:

According to the description, if we take a line perpendicular (meaning under an angle of 90 degrees, right?) to any root, then the whole space should somehow reflect on that line, which I don´t understand.

Then, this is also the reason why I don´t understand Weyl group, but let´s leave this to a separate question.

Thank you for any help.

Answer 1

Edit: I realized I have not addressed your actual question, which was concerend with the first criterion there, that the root vectors span the space $E$. You asked:

So does that mean that these root vectors have linear span or are something like a basis?

It cannot mean the first: Because remember that every (any) set of vectors span some vector space, namely their, well, span. So that would not be a condition at all.

It does not mean that the roots form a basis: Look at the example of the root system $A_2$ (six root vectors which point to the vertices of a perfect hexagon). They all lie in a plane of just two dimensions, so not all six vectors together are a basis.

What it literally means is that the root vectors are a generating system for their vector space $E$. They can fail to be a basis because they need not all be linearly independent. But they need to span their space (not just all lie in some lower-dimensional subspace).

Maybe it helps to see how this mild condition could possibly fail. Which of these examples a-g are root systems?

$a. \Phi = \{(1,0), (-1,0)\}, E= \mathbb R^2 \\ b. \Phi = \{(1,0), (-1,0)\}, E= \{ v=(v_1,v_2) \in \mathbb R^2: v_2 =0\}\\ c. \Phi = \{(1,0,-1),(-1,0,1),(1,-1,0),(-1,1,0),(0,1,-1),(0,-1,1)\}, E= \mathbb R^3 \\ d. \Phi= \{(1,0,-1),(-1,0,1),(1,-1,0),(-1,1,0),(0,1,-1),(0,-1,1)\}, E= \{v= (v_1,v_2, v_3) \in \mathbb R^3: v_1+v_2+v_3=0 \} \\ e. \Phi= \{(1,0),(-1,0),(-1/2,\sqrt3/2),(1/2,-\sqrt3/2),(1/2,\sqrt3/2)(-1/2,-\sqrt3/2)\}, E= \mathbb R^2 \\ f. \Phi = (0,-1.383,17,\sqrt\pi), (0,1.383,-17,-\sqrt \pi), E= \mathbb R^4 \\ g. \Phi = (0,-1.383,17,\sqrt\pi), (0,1.383,-17,-\sqrt \pi), E= span \langle(0,-1.383,17,\sqrt\pi)\rangle = \{(c,-1.383c,17c,\sqrt\pi c) \in \mathbb R^4: c \in \mathbb R\}$

Answer: a, c, and f are not because they fail criterion 1. In these examples, the space $E$ which we specified is "too big", it is not generated by the roots in $\Phi$ (in examples a and f, the roots in $\Phi$ span only a $1$-dimensional line and do not generate the spaces $E$ which are $2$- and $4$-dimensional, respectively; in example c, the span of $\Phi$ would be a two-dimensional plane, and does not span the entire vector space $E$ which has dimension $3$.

In all other examples, indeed $\Phi$ spans $E$ (and you will see that b,d, and g, are basically just "corrected" versions of a,c, and f, where we now chose $E$ "smaller" so that it is the span of $\Phi$). Example e is exactly the way you see the root system $A_2$ in the plane, note that here these vectors obviously do span the full $E= \mathbb R^2$.

Finally, we have not checked it yet, but indeed examples b, d, e, and g also satisfy all the other criteria, so indeed they are root systems. However, it should be very easy to see that a and g are actually "the same" (isomorphic). Indeed, they are two manifestations of the root system $A_1$ ; in a way, just different parametrizations of the same idea: It's just some non-zero vector and its negative, sitting in the line it spans.

It is not quite that easy, but a very good exercise to see that examples d and e are also isomorphic to each other. They are different parametrizations (i.e. choice of underlying coordinates) of the root system $A_2$ which you see in the image there and which I will talk about in the rest of this answer. Think for yourself: In example e, which two of the roots could be $\alpha$ and $\beta$ in the picture? And what pairs of the six vectors in example d could be $\alpha$ and $\beta$?

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A root system is just a bunch of vectors (best visualized in plain old Euclidean space) that has, in a sense made precise by the axioms, a lot of symmetry.

The basic one among these symmetries goes like this: Take any of the roots. Imagine the (hyper)plane which stands perpendicular on this root. Reflect everything at that hyperplane (i.e. use that plane as a mirror). Then all your roots have been reflected to another root, i.e. the whole thing thus reflected looks exactly as before.

Those six vectors which you see in the picture you uploaded are one example of a root system. See how their endpoints form a perfect hexagon. See how the angle between each root and its two neighbours on the left and right is exactly $60^°$. Imagine a coordinate system behind it, with the centre the origin, and all roots of length $1$.

Let's check those reflections: First, take the root called $\alpha$. What is the "hyperplane perpendicular to it"? We have just two dimensions in this example, so a plane is just a line; indeed, here it's just the vertical axis, which quite visibly is made up of everything that stands perpendicular to $\alpha$. OK, so let's reflect everything at that axis. What happens? Each of the roots finds its perfect mirror image: $\alpha$ switches with $-\alpha$, $\beta$ with $\alpha+\beta$, and $-\beta$ with $-\alpha-\beta$. Cool.

But what if instead, for example, we start with the root $\beta$. Then what is the "hyperplane perpendicular to it"? Again it's just a line, but you have to imagine it, it is not drawn in the picture: It passes through the origin at an angle of $30^°$ above the root $\alpha$ on the horizontal axis, i.e. it cuts through exactly halfway between $\alpha$ and $\alpha+\beta$ in the first quadrant, and halfway between $-\alpha$ and $-\alpha-\beta$ in the third quadrant. OK, so let's use that one as mirror of the whole thing. What happens? Again, each root finds a perfect mirror image. This time, the reflection switches $\alpha \leftrightarrow \alpha +\beta$, $\beta \leftrightarrow -\beta$, and $-\alpha \leftrightarrow -\alpha-\beta$.

I leave it to you to imagine the reflection at the line perpendicular to the root $\alpha+\beta$.

Now in this root system, which as said in a way is a perfect hexagon, these things work. But now draw a proposed root sytem whose endpoints are a perfect pentagon. Try the reflections. You will see that some of them do not work: Some roots, when reflected, do not "land on" other roots. So those vectors (whose endpoints form a perfect pentagon) do not form a root system.

You should see that most other perfect $n$-gons do not work either! As a rare exception beside the hexagon, four vectors whose endpoints form a square do work; but, three vectors whose endpoints form a perfect triangle do not (when you try the reflections, you will see you have to "complete" the triangle to a hexagon like above).

Now beyond this basic one, there is a little more symmetry needed to make something a root system, which rules out even more vector sets even though on first sight they might look pretty symmetric.

For example, it turns out that in any root system, any two of the roots are only "allowed" to have either an angle of $0, 30, 45, 60, 90, 120, 135, 150$, or $180$ degrees between them. It also turns out that "many" of the roots have to have the same length. Etc. (In the above example, all roots have the same length, and you see that the possible angles between any two roots are only $0, 60, 120$, or $180$ degrees.)

This should be enough for a very coarse first idea. I advertise my related answers to How to visualize intuitively root systems and Weyl group?, and, already on an advanced level, Picture of Root System of $\mathfrak{sl}_{3}(\mathbb{C})$.

It's a whole different question why we are interested in root systems! The short answer to that is that if we know how to translate from them, they tell us a lot about the structure of certain groups and Lie algebras. However, this translation is quite involved and passes thorugh several levels of abstraction. One does not "get it" if one is not willing to cross-check a lot of things in examples and exercises involving many matrix computations. Some examples of this "translation" happen in my answers to Basic question regarding roots from a Lie algebra in $\mathbb{R}^2$ and Understanding an eigenspace of semi-simple elements in Lie algebra.

Answer 2

So, per the definition(I think I heard the term before though), a root system IS a span of vectors and their additive inverses and has the property that any of their projections on to any other is a rational multiple(in this case limited to $\mathbb{Z}/2$ but irrelevant for discussion) of the other and the vectors have nice symmetry properties.

It's saying we have several vectors from E that we want to treat as a special collection, but for the collection to behave as we want we need want it to have these properties(that we think are special).

Now root systems sound like taking a root which is the inverse of a power: $\sqrt{x} = y \Leftrightarrow x = y^2$. Requiring the stuff about rationality seems to have something to do with solving rational equations or what is essentially the same as solutions of algebraic polynomials. I say this because it feels a lot like how Galois theory proceeds down the rational transitivity train of thought.

1. The root system $\Phi$ span E. This tells us we can write any vector in E as a linear combination of vectors in $\Phi$. [This is related to representation theory, we are writing other vectors in special formats to hopefully be able to extract useful information]

2. The only scalar multiples in $\Phi$ are the reflected vectors. Hence, remove the reflected vectors and you have no scalar multiples and hence a basis. If B is a basis for E then {B, -B} is a "root system" so far.

3. This is taking 2 and trimming the fat. It's not just any set B and it's -B BUT they have to all have this reflection property. What is this reflection property? Look at the diagram and reflect all the vectors across some perp to some vector. You get symmetry! That is what they are requiring with this constraint. They don't want just any vectors(which would be far more random) but one that basically has reflective symmetry at every dimension of the span. One then is dealing in group theory because one can talk about permuting the vectors and such. Reflections will be automorphisms.

4. We additionally require that these vectors in $\Phi$ not just be relatively arbitrarily chosen(satisfying 1 & 2) BUT also that any projection of one on to another can only give a rational alteration. Hence all the projections of each $v \in \Phi$ will be rationally related. In the specific definition it says half integers. I believe this only to be a simplification "trick" but even if not it just severely limits the choices for the vectors positions. b would have to project to either a, a/2, or -a, or -a/2, with c it's either c, -c, c/2, or -c/2 or any multiples. (so it's sorta like harmonics, integral multiples but we also can have half-multiples). It just limits the hyper-hedron's structure to be even more regular. I think representationally it changes nothing but coding. E.g., similar to number bases. They are only for representation but are all equivalent in their ability to represent. I suppose depending on what these are used for it could matter.

3 & 4 is the big property that really makes this root system interesting. One constrains the "basis"(a span but lets think of it more was a "group" as it's sort of it's own little "vector" space that we can do things in and then write everything else in terms of) to have this symmetry property and the other constrains it to have these rational relationships. This is effectively trying to represent group and number theory in a more "vectorized" format which might then realize some new insights in to problems.

Ultimately, 3 & 4 just create a "configurational structure" out of the vectors. That structure is analogous to regular geometric figures. The idea, as far as just getting a feel for the concept/definition is to think of in general terms about what it's trying to achieve/create even if that means generalizing parts of it(this much of what mathematics is in any case). Once you have the vague idea of the kind of structure being created(a definition is a template or blueprint) then you can start to fill in the specifics. Pointless to try to understand it from a detailed view when you do not even understand the undetailed view.

Your main issue, as you say, is 3. It's best to take, say, 2 vectors as a basis for R^2 and then add in their reflects and then study (3).

E.g.,

$a = (1,0), b = (0,1), \Phi = \{a,-a,b,-b\}$

Then the plane perpendicular to $a$ is the plane given in which $a$ is the normal vector. This plane has both $b$ and $-b$ in it and so the reflection works.

Take $b = (0.5,1)$. Now we have $\Phi = \{a,-a,b,-b\}$ BUT when we reflect all the vectors now(b and -b) we end up with new vectors that are not in $\Phi$. In this case $\Phi = \{a,-a,b,-b,(-0.5,1),(-0.5,-1)\}$. See the difference? We had to actually add 2 more vectors to make sure the reflection(mirror reflection in some sense) was valid.

You could think of building up a basis using these methods. Start with a single vector a. You have to add -a. Ok, Now you want to add another vector b? Well, not just any b will work. You have to project it on to a and make sure they are in rational proportion. Once you do that you have to add -b. But also because we have to reflect b and -b across the plane normal to a we will get vectors b' and -b'(b' could be b). But also we will get a' and -a'(potentially). Choose a new vector c and repeat this process.

By understanding it from the bottom up and realizing how to construct such sets(similar to how Graham Schmidt is used) it will make more sense how they work. What we end up with is a structure that has complete reflective symmetry in all dimensions. It can be thought of as a "hyperdimensional mirror". The number of vectors are sort of like the vertices of a hyper-dimensional simplex that has nice symmetry(but can be quite complex) and relationships between the lengths of the edges that connect them.

It's quite interesting, I'll have to study it a bit more to try and see what use it is. Off to the wiki page I go to get lost ;)

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I will construct a bi-integral R^2 root system from scratch following the lines of Graham-Schmid since there seems to be some confusion. I will actually do this first for a rational root system then specialize to get bi-integral.

let $a = (1,0)$ be our starting vector. In our mock root system we currently have some element a. We will simply construct all possible rational root systems based on $a$ and hence all others will be isomorphic(rotational symmetry). We will call this set R and update it as we go along: R = {a, -a}. -a must be in R according to C2 and/or C3. (Cn is the nth criteria)

We wish to find $b_1 = (x,y)$ such that $a$ and $b_1$ are compatible with being in a root system. Clearly $b_1 = \pm a$ is useless.

According to C4 we must have that the projection of $b$ on to $a$ has a special relationship to $a$'s length.

The projection of $b$ onto $a$ is given by $b^{\rightarrow}a = \frac{b^{T}a}{a^T a} a = x \hat{a}$ and $\frac{||b^{\rightarrow}a||}{||a||} = x$. C4 requires $x = \frac{k}{2}$(equivalently $x \in \mathbb{Z}/2$) or more generally $x \in \mathbb{Q}$.

$b$ is a vector that satisfies C4. Now we add it to our general root system $R := \{a,-a,b\}$ which we also now have to add $-b$ according to C2(and C3 will require this too). So $R := \{a,-a,b,-b\}$.

It should be obvious at this point that we have a span and $a$ and $b$ form a basis for $\mathbb{R}^2$(as does any absolutely different vectors from R). It is obvious that when we chose $b$, we required it not to be parallel to $a$ and this forced independence with a, in generally with more vectors in the root system this may not be the case.

Now at this stage we have to carry out C3 on all our vectors so we actually get a root system. We might end up with new vectors who then have to be specially created to fit.

When we mirror the vector $b$ along the hyperplane normal to $a$ this is just a mirror around $(0,1)$. C3 then requires that the reflected vector of b also be in R. This reflection is simply flipping the x component. The $-a$ flipped both. The $-a$ flipped both coordinates giving us the reflection for the vector itself(this will be true from C3 so C2 is sort of redundant). But for $b$ and $-b$ this will flip only the first component(generally it is going to be a rotation mirror and then inverse rotation).

So to satisfy C3 we have to add $b' = (-x, y)$ and this gets us $-b'$ too. $R := \{a,-a,b,-b,b',-b'\}$

But when we added all these new vectors we also have to reflect around every single one, a was not special. We could have started out with $b$ first then arrived at $a$. In doing this we would then get $R := \{a,-a,b,-b,b',-b',a',-a'\}$.

We have to continue to apply C3 and C4(e.g., similar to GS in the abstract). We will end up with many "$x$"'s on for each vector(I didn't parameterize them for convenience but hopefully you get the picture).

Carrying this out will produce a root system with potentially an infinite number of vectors depending on the number theoretic choices of our "x"s and "y"'s. By choosing only some $half-integer multiple$ for our $x's$ and possibly the $y's$ we can get a finite process that will converge. E.g., the standard basis will produce such a system as long as we allow 0 to be a valid half-integer multiple. The standard root system is $\mathbb{R}^n$ = the standard basis except there is quite a bit of redundancy. The minimal spanning standard root system = the standard basis for $\mathbb{R}^n$.

If we add a compatible vector to $R$ then we will have new vectors that are created by C3. C3 will exponentially generate, in most cases, all the "house of mirror" reflections that take place. By requiring the more limiting $bi-integrality$ condition one reduces the complexity significantly of the structure being created.

For $\mathbb{R}^3$ the process works exactly the same and since $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3$ we can see that the complexity even of a 2D root system just gets exponentiated when moving to higher dimensions, but still is fundamentally the same simple process. These root system are very fractal like due to C3 generating these new vectors.