# I Relation Between Cross Product and Infinitesimal Rotations

#### dm4b

Looking into the infinitesimal view of rotations from Lie, I noticed that the vector cross product can be written in terms of the generators of the rotation group SO(3). For example:

$$\vec{\mathbf{A}} \times \vec{\mathbf{B}} = (A^T \cdot J_x \cdot B) \>\> \hat{i} + (A^T \cdot J_y \cdot B) \>\> \hat{j} + (A^T \cdot J_z \cdot B) \>\> \hat{k}$$

where,

$$J_x = \begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix} \>\>\>;\>\>\>;J_y = \begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix} \>\>\>;\>\>\>;J_z = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$$

I couldn't help but think I was missing some profound connection here. Why is the vector cross product seemingly intimately related to the generators of the SO(3) rotation group?

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#### fresh_42

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I couldn't help but think I was missing some profound connection here. Why is the vector cross product seemingly intimately related to the generators of the SO(3) rotation group?
Because both are the same three dimensional simple Lie algebra - up to isomorphism.

#### dm4b

Because both are the same three dimensional simple Lie algebra - up to isomorphism.
Hello, I’m not completely appreciating your comment. Can you provide more detail how the vector cross product relates to the Lie algebra?

#### dm4b

Because both are the same three dimensional simple Lie algebra - up to isomorphism.
Or, maybe I should phrase my question more simply as ... what does the vector cross product have to do with rotations?

#### fresh_42

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Or, maybe I should phrase my question more simply as ... what does the vector cross product have to do with rotations?
Not the rotations, the infinitesimals of rotations, the tangent vectors. There is only one three dimensional real Lie algebra without proper ideals, $\mathfrak{sl}(2)$. The versions $\mathfrak{so}(3)$ and $\mathfrak{su}_\mathbb{R}(2)$ are isomorphic copies. Here are the basis transformations:
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
$\{U,V,W\}$ is the cross product, $\{H,X,Y\}$ is the special linear Lie algebra, $\{\sigma_k\}$ the Pauli matrices, i.e. $\{i\cdot \sigma_k\}$ is the unitary version, and in part one of the article you can see the connection between the unitary group and the orthogonal group.
Hello, I’m not completely appreciating your comment. Can you provide more detail how the vector cross product relates to the Lie algebra?
The cross product $U\times V = W$ can be thought of as the Lie product $[U,V]=W$.

• Cryo and dm4b

#### dm4b

Not the rotations, the infinitesimals of rotations, the tangent vectors. There is only one three dimensional real Lie algebra without proper ideals, $\mathfrak{sl}(2)$. The versions $\mathfrak{so}(3)$ and $\mathfrak{su}_\mathbb{R}(2)$ are isomorphic copies. Here are the basis transformations:
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
$\{U,V,W\}$ is the cross product, $\{H,X,Y\}$ is the special linear Lie algebra, $\{\sigma_k\}$ the Pauli matrices, i.e. $\{i\cdot \sigma_k\}$ is the unitary version, and in part one of the article you can see the connection between the unitary group and the orthogonal group.

The cross product $U\times V = W$ can be thought of as the Lie product $[U,V]=W$.
Thanks, I will check out that article. I am better appreciating the connection, but still struggling with the following.

I work in the aerospace industry, specifically we deal with modeling and simulation environments for aerospace flight vehicles. Quite often those generators (the matrices) are seen associated with the implementation of a vector cross product in the sims. However, the fact that these are generators of the rotation group largely goes unnoticed. Thus, I am trying to figure out how to explain this connection in "laymen's terms" for non-mathematicians.

I hark back to Feynman who I believe said something like, "if you cannot explain your theory to the layman, there is either something wrong with you or your theory". Pretty sure the theory is good here, so I am still feeling like there is something wrong with me, heh

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#### fresh_42

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Let me put it this way:
The rotation matrix around the $x-$axis is $\begin{bmatrix}1&0&0\\0&\cos \varphi &-\sin \varphi \\ 0 &\sin \varphi & \cos \varphi \end{bmatrix}$ and its tangent is $\begin{bmatrix}0&0&0\\0&-\sin \varphi &-\cos \varphi \\ 0 &\cos \varphi & -\sin \varphi \end{bmatrix}$ which at $0°$ gets $\begin{bmatrix}0&0&0\\0&0 &-1 \\ 0 &1 & 0 \end{bmatrix}$. This is your matrix $J_x$ (in which I think you confused the row/column numbers).

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#### dm4b

Thanks, this also helps. (Fixed the typo in the OP, as well). Gonna read up on all this some more and give it some thought - it's slowly all coming together.

#### fresh_42

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Thanks, this also helps. (Fixed the typo in the OP, as well). Gonna read up on all this some more and give it some thought - it's slowly all coming together.
I had also a typo to fix: It's the tangents at $0^°$, which makes sense, as the Lie algebra of a Lie group are those tangents at the identity element $1$ of the group which becomes the $0$ of the tangent vector space aka Lie algebra.

#### samalkhaiat

Science Advisor
Why is the vector cross product seemingly intimately related to the generators of the SO(3) rotation group?
This is because the structure constant of $SO(3)$ (that is $\epsilon_{abc}$) also appears in the definition of the cross product on $\mathbb{R}^{3}$:

$$\big[ T_{a} , T_{b} \big] = \epsilon_{cab} \ T_{c} ,$$ $$( \vec{A} \times \vec{B} )_{c} = \epsilon_{cab} A_{a}B_{b} = A_{a}(J_{c})_{ab}B_{b},$$
where $(J_{c})_{ab} = \epsilon_{cab}$ is the $ab$ matrix element of the generators in the adjoint (=fundamental) representation of $SO(3)$. For the same reason, you can convert the cross product on $\mathbb{R}^{3}$ into commutator of (Lie algebra valued) matrices: Define the matrices $\mathbb{A} = T_{a}A_{a}$ and $\mathbb{B} = T_{a}B_{a}$ in some arbitrary representation $\{T_{a}\}$ of $so(3)$. Also, write $\vec{C} = \vec{A} \times \vec{B}$. Now, rewrite the cross product as $$C_{a} = \epsilon_{abc} A_{b}B_{c}.$$ If you contract both sides with $T_{a}$ and use the algebra $\big[ T_{a} , T_{b} \big] = \epsilon_{abc}T_{c}$, you obtain $$\big[ \mathbb{A} , \mathbb{B} \big] = C_{a}T_{a} \equiv \mathbb{C}.$$
In fact, the concept of cross product can be defined for any compact Lie algebra using the (real and totally anti-symmetric) structure constant: $$( \vec{V} \times \vec{U})_{a} = C_{abc} V_{b}U_{c} ,$$ and the above case of the algebra $so(3)$ is just the special case when $C_{abc} = \epsilon_{abc}$.

• dm4b

#### fresh_42

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In fact, the concept of cross product can be defined for any compact Lie algebra
Lie algebras $\mathfrak{g}$ are not compact: $X \in \mathfrak{g} \Longrightarrow c\cdot X \in \mathfrak{g}$ for all $c \in \mathbb{R}$.
For my part, I will not try to find a finite subcover.

Lie groups as topological groups can be compact, but not their Lie algebras. Lie algebras are usually not regarded as topological vector spaces, and if, they are isomorphic to $\mathbb{R}^n$ or $\mathbb{C}^n$ - other fields apart.

#### samalkhaiat

Science Advisor
Lie algebras $\mathfrak{g}$ are not compact: $X \in \mathfrak{g} \Longrightarrow c\cdot X \in \mathfrak{g}$ for all $c \in \mathbb{R}$.
For my part, I will not try to find a finite subcover.

Lie groups as topological groups can be compact, but not their Lie algebras. Lie algebras are usually not regarded as topological vector spaces, and if, they are isomorphic to $\mathbb{R}^n$ or $\mathbb{C}^n$ - other fields apart.
The Cartan metric $g_{ab} = - C^{c}{}_{ae}C^{e}{}_{bc}$ (which, in the adjoint representation, is equal to $\mbox{Tr}(J_{a}J_{b})$ ) can be used to define an inner product on the Lie algebra.
Using the fact that the adjoint representation $A(x) = \exp a(x)$ of a Lie group $G$ is orthogonal if and only if the adjoint group $G/Z$ is compact, the compactness of a Lie group can be expressed in terms of its Lie algebra.
A simple or semi-simple Lie algebra is said to be compact if the Cartan metric $g_{ab}$ is positive-definite.
We also have the structure theorem of compact Lie algebras: Any compact Lie algebra is the direct sum of irreducible compact Lie algebras, where irreducible means either simple or one-dimensional.
See Cambridge monograph: “Group structure of gauge theories”, L. O’RAIFEARTAIGH., Section 3.4: Compact Lie algebras.

#### fresh_42

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The Cartan metric $g_{ab} = - C^{c}{}_{ae}C^{e}{}_{bc}$ (which, in the adjoint representation, is equal to $\mbox{Tr}(J_{a}J_{b})$ ) can be used to define an inner product on the Lie algebra.
It is an inner product on the weight space, not the Lie algebra! The inner product is defined on the $\mathbb{Q}-$linear hull of the roots, i.e. on $CSA^*$ not the Lie algebra vectors.
Using the fact that the adjoint representation $A(x) = \exp a(x)$ of a Lie group $G$ is orthogonal if and only if the adjoint group $G/Z$ is compact, the compactness of a Lie group can be expressed
in terms of its Lie algebra.
A simple or semi-simple Lie algebra is said to be compact if the Cartan metric $g_{ab}$ is positive-definite.
We also have the structure theorem of compact Lie algebras: Any compact Lie algebra is the direct sum of irreducible compact Lie algebras, where irreducible means either simple or one-dimensional.
See Cambridge monograph: “Group structure of gauge theories”, L. O’RAIFEARTAIGH., Section 3.4: Compact Lie algebras.
I think I have found the source of confusion:

What does exist are compact real forms, and a real semisimple Lie algebra $\mathfrak{g}$ is said to be of compact type iff its adjoint group is a compact subgroup of the general linear group $GL(\mathfrak{g})$. This definition btw. is again via the compactness of a group (sic!) not a vector space.
$\omega\, : \,X \longmapsto - \operatorname{tr}(\operatorname{ad}(X))^2$ is a positive definite quadratic form (Casimir polynomial) for semisimple Lie algebras, if $G$ is a compact group. It has a corresponding bilinear form given as $\beta(X,Y) = \omega(X+Y)-\omega(X)-\omega(Y)$.

For $\mathfrak{sl}(2)$ with the basis $\{\,H,X,Y\, : \,[H,X]=2X\, , \,[H,Y]=-2Y\, , \,[X,Y]=H\,\}$ I got $\omega(h,x,y) = 8h^2+8xy$ with the corresponding matrix $\begin{bmatrix}8&0&0\\0&0&4\\0&4&0\end{bmatrix}$ which has eigenvalues $8,2,-2$ and is thus not definite. The crucial point is again, that although there is only one three dimensional simple Lie algebra, this doesn't mean the Lie groups are the same.

As a (complex) semisimple Lie algebra admits a real form of compact type, I guess another basis is necessary but I didn't went through the proof. It suggests a orthogonal group to work with. Maybe O’Raifeartaigh is more sloppy than Varadarajan and calls Lie algebras of compact type (quadratic form, adjoint group is compact) compact themselves. However, I think this is very misleading, since a Lie algebra is still a Euclidean space! The terminology of compactness in this context goes always over a topological group, which does allow compactness. To call $\mathbb{C}^n$ compact is quite a stretch in my opinion!

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#### bolbteppa

Given a function $f$ of a vector $\mathbf{r}$, $f(\mathbf{r})$, the function can also be considered with the rotated vector as the argument of $f$ using a rotation matrix $R$, $f(R\mathbf{r})$. An infinitesimal rotation of a vector $\mathbf{r}$ can be written abstractly as $\mathbf{r} + \delta \mathbf{r}$ so that $f$ expands infinitesimally as $f(\mathbf{r} + \delta \mathbf{r}) = f(\mathbf{r}) + \delta \mathbf{r} \cdot \nabla f$, where $\delta \mathbf{r}$ can be explicitly found for an infinitesimal rotation by visualizing it as a vector perpendicular to the plane spanned by $\mathbf{r}$ and a vector along the axis of rotation whose length is the magnitude of the rotation, so that $\delta \mathbf{r} = \delta \vec{\theta} \times \mathbf{r}$, so that (using $A \cdot B \times C = C \cdot A \times B = ..$)

$f(\mathbf{r} + \delta \mathbf{r}) = f(\mathbf{r}) + \delta \mathbf{r} \cdot \nabla f= f(\mathbf{r}) + \delta \vec{\theta} \times \mathbf{r} \cdot \nabla f = f(\mathbf{r}) + \delta \vec{\theta} \cdot (\mathbf{r} \times \nabla ) f$

which shows that the infinitesimal rotation of a vector is determined by $\mathbf{r} \times \nabla$ which is the cross product of two vectors $\mathbf{r}$ and $\nabla$, so any cross product should be related to the infinitesimal rotation matrices one find by expanding the action of a rotation matrix $R$ on a vector $\mathbf{r}$ infinitesimally in matrix form, $R(\mathbf{r}) = (I + \delta \vec{\theta} \cdot \vec{J})\mathbf{r}$.

Whether we begin by using functions of vectors, or working with rotation operators acting on vectors, ultimately we are analyzing the infinitesimal structure of the rotation group, a continuous group, and so one can abstract to working with the infinitesimal structure of continuous groups called Lie groups, the infinitesimal structure being known as a Lie algebra, the above matrix or differential operator forms being representations of the Lie algebra of the rotation group, and thus relate it to the structure theory of Lie algebras as in this post.

#### samalkhaiat

Science Advisor
It is an inner product on the weight space, not the Lie algebra! The inner product is defined on the $\mathbb{Q}-$linear hull of the roots, i.e. on $CSA^*$ not the Lie algebra vectors.

I think I have found the source of confusion:

What does exist are compact real forms, and a real semisimple Lie algebra $\mathfrak{g}$ is said to be of compact type iff its adjoint group is a compact subgroup of the general linear group $GL(\mathfrak{g})$. This definition btw. is again via the compactness of a group (sic!) not a vector space.
$\omega\, : \,X \longmapsto - \operatorname{tr}(\operatorname{ad}(X))^2$ is a positive definite quadratic form (Casimir polynomial) for semisimple Lie algebras, if $G$ is a compact group. It has a corresponding bilinear form given as $\beta(X,Y) = \omega(X+Y)-\omega(X)-\omega(Y)$.

For $\mathfrak{sl}(2)$ with the basis $\{\,H,X,Y\, : \,[H,X]=2X\, , \,[H,Y]=-2Y\, , \,[X,Y]=H\,\}$ I got $\omega(h,x,y) = 8h^2+8xy$ with the corresponding matrix $\begin{bmatrix}8&0&0\\0&0&4\\0&4&0\end{bmatrix}$ which has eigenvalues $8,2,-2$ and is thus not definite. The crucial point is again, that although there is only one three dimensional simple Lie algebra, this doesn't mean the Lie groups are the same.

As a (complex) semisimple Lie algebra admits a real form of compact type, I guess another basis is necessary but I didn't went through the proof. It suggests a orthogonal group to work with. Maybe O’Raifeartaigh is more sloppy than Varadarajan and calls Lie algebras of compact type (quadratic form, adjoint group is compact) compact themselves. However, I think this is very misleading, since a Lie algebra is still a Euclidean space! The terminology of compactness in this context goes always over a topological group, which does allow compactness. To call $\mathbb{C}^n$ compact is quite a stretch in my opinion!
If $G$ is compact, then the real vector space $\mathfrak{g}$ admits a $\mbox{Ad}(G)$-invariant, positive-definite inner product
$$\langle X , Y \rangle = \int_{G/Z} d \mu (g) \left( \mbox{Ad}(g)X , \mbox{Ad}(g)Y \right) ,$$ where $X = x^{a}J_{a}, \ Y = y^{a}J_{a}$ and $\left( X , Y \right) = g_{ab}x^{a}y^{b}$. With respect to the above inner product, the elements of $\mbox{Ad}(G)$ act by orthogonal transformations, and the elements of $\mbox{ad}\mathfrak{g}$ act by anti-symmetric transformations. The proof can be found in L. O’Raifeartaigh or Anthony W. Knapp (2002) “Lie Groups Beyond an Introduction”, Progress In Mathematics, 140. Proposition 4.24.

By the way, I chose O’Raifeartaigh for you because he is not “sloppy”. You can, if you wish, look up Knapp’s where the properties of compact Lie algebras are explained reasonably well.

#### fresh_42

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If $G$ is compact, then the real vector space $\mathfrak{g}$ admits a $\mbox{Ad}(G)$-invariant, positive-definite inner product
$$\langle X , Y \rangle = \int_{G/Z} d \mu (g) \left( \mbox{Ad}(g)X , \mbox{Ad}(g)Y \right) ,$$ where $X = x^{a}J_{a}, \ Y = y^{a}J_{a}$ and $\left( X , Y \right) = g_{ab}x^{a}y^{b}$. With respect to the above inner product, the elements of $\mbox{Ad}(G)$ act by orthogonal transformations, and the elements of $\mbox{ad}\mathfrak{g}$ act by anti-symmetric transformations. The proof can be found in L. O’Raifeartaigh or Anthony W. Knapp (2002) “Lie Groups Beyond an Introduction”, Progress In Mathematics, 140. Proposition 4.24.

By the way, I chose O’Raifeartaigh for you because he is not “sloppy”. You can, if you wish, look up Knapp’s where the properties of compact Lie algebras are explained reasonably well.
Agreed.

And how does it make $\mathfrak{g} \cong \mathbb{R}^n$ compact in its usual topological sense?7
for any compact Lie algebra

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#### samalkhaiat

Science Advisor
In #13, you said
It is an inner product on the weight space, not the Lie algebra!
So, the proposition in #15 was to tell you that the matrix $g_{ab}$ can be used to define an inner product on the Lie algebra of a compact Lie group. This proposition, as I have already stated in #12, allows us to address the compactness of a Lie group in terms of its Lie algebra. This does not mean that the vector space $\mathfrak{g} \cong \mathbb{R}^{n}$ is compact. $\mathbb{R}^{n}$ is a locally compact topological group under vector addition.
Again in #12, I defined the term “compact Lie algebra” by
A simple or semi-simple Lie algebra is said to be compact if the Cartan metric $g_{ab}$ is positive-definite.
Or, a compact Lie algebra consists of the (infinitesimal) generators of a compact Lie group, i.e., a compact Lie algebra is the Lie algebra of a compact Lie group.
The above two definitions are exactly the definitions given by mathematicians to the term “compact Lie algebra”.
Example (here I will use physicists notation for the algebra, $\big[T_{a} , T_{b} \big] = i C_{ab}{}^{c}T_{c}$): The Lie algebra $\mathfrak{so}(1,2)$ of the non-compact Lorentz group $SO(1,2)$ is an example of simple Lie algebra that is not compact:
$$\big[T_{1} , T_{2} \big] = - i T_{3}, \ \ \big[T_{2} , T_{3} \big] = i T_{1}, \ \ \big[ T_{3} , T_{1} \big] = i T_{2} .$$
Clearly, structure constant is real but not totally anti-symmetric, for example $C_{12}{}^{3} = C_{13}{}^{2} = -1$. Also the metric is diagonal, $g_{11} = g_{22} = -g_{33} = -2$, but not positive.
However, any simple Lie algebra can be put into a compact form if we allow for complex linear transformations of the generators (which leads to a different Lie group). For instance, in the above example we only need to define the following set of generators: $t_{1} = iT_{1}, \ t_{2} = iT_{2}, \ t_{3} = T_{3}$. In terms of the set $\{t_{a}\}$, the above commutation relations become
$$[t_{1} , t_{2}] = it_{3}, \ \ [t_{2} , t_{3}] = it_{1}, \ \ [t_{3} , t_{1}] = it_{2}.$$
The structure constant is now real and totally anti-symmetric $C_{ab}{}^{c} = \epsilon_{abc}$, and $g_{ab} = 2 \delta_{ab}$. So, according to our definition, the algebra of the set $\{t_{a}\}$ is compact. Indeed, it is nothing but the real Lie algebra $\mathfrak{so}(3)$ of the compact group $SO(3)$ of rotations in $\mathbb{R}^{3}$.

In our trade, we approach the subject as follow: we would like to write a Lagrangian for the non-abelian gauge fields. Lorentz-invariance suggests the form $$\mathcal{L} \sim g_{ab} F^{a}_{\mu\nu}F^{b\mu\nu} ,$$ with $g_{ab}$ being a constant (may be taken symmetric) matrix but It must be real so that the Lagrangian is real. Next, canonical quantization and the positivity of the quantum mechanical scalar product restrict $g_{ab}$ to be positive-definite. And finally, the gauge invariance of $\mathcal{L}$ imposes the following condition on $g_{ab}$ $$g_{ae}C_{bc}{}^{e} + g_{be}C_{ac}{}^{e} = 0. \ \ \ (1)$$ This means that, for $\mathcal{L}$ of the above form, the gauge group must be compact.
And we summarize the above by the following three equivalent conditions:
I) There exists a real positive-definite matrix $g_{ab} = g_{ba}$ that satisfies the gauge invariance condition (1).
II) There is a basis for the Lie algebra (i.e., a real non-singular linear transformation $\bar{T}_{a} = M_{ab}T_{b}$) for which the structure constant $\bar{C}_{ab}{}^{c}$ is anti-symmetric in an all three indices. In this basis we simply write $\bar{C}_{abc}$ instead of $\bar{C}_{ab}{}^{c}$.
III) The Lie algebra is the direct sum of commuting compact simple and $\mathfrak{u}(1)$ subalgebras. Correspondingly, the gauge group is the direct product of simple compact Lie groups and one-parameter compact groups $U(1)$, modulo a discrete centre.

#### fresh_42

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I don't mind you to type an entire book on Lie algebras here, but this does not make
Again in #12, I defined the term “compact Lie algebra” by
A simple or semi-simple Lie algebra is said to be compact if the Cartan metric $g_{ab}$ is positive-definite.
less nonsense. It is wrong and a misinformation!

I assume that O’Raifeartaigh calls it as such, but this is sloppy and misleading. A Lie algebra is still isomorphic to say $\mathbb{R}^n$ and an inner product does not make it compact. It suggests that a Euclidean space is compact if you only find a suitable inner product. Sorry, but this is nothing less than wrong. I do not doubt that the Killing-form is non-degenerate and that there is a complex basis in which it is positive definite, but again, this does not make $\mathbb{R}^n$ compact. I'm not objecting the theorems, I fight against the term compact in this context.

Please stop spreading this false wording! Varadarajan phrases it as "$\mathfrak{g}$ admits a compact form" which is not the same as to say it is compact. Compactness is a precisely defined topological term, and to abuse it confuses people. It obviously confuses you, so how much more will it confuse students?

As long as you cannot give me finite subcover of
$$\mathfrak{sl}(2,\mathbb{C}) \subseteq \bigcup_{n\in \mathbb{N}} \left\{ \begin{bmatrix}a&b\\c&-a\end{bmatrix}\,:\, (a,b,c)\in n \cdot \{\,z\in \mathbb{C}\,:\,||z||<1\,\}\right\}$$
in which you may change the norm accordingly, but not the covering, as long is compact in this context simply wrong.

#### bolbteppa

It is not false, you are misunderstanding the meaning of the word compact in the context of Lie algebras - this use of compact in Lie algebras is very common, e.g. Barut's Group Representations Ch. 1 defines compactness of a Lie algebra by saying a Lie algebra is compact if there exists a positive-definite quadratic form satisfying $([X,Y],Z) = (X,[Y,Z])$, and the definition comes from (as Barut points out in a footnote) the fact that the Lie algebra of a compact Lie group (in the topological sense) is compact in this sense.

#### fresh_42

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It is not false, you are misunderstanding the meaning of the word compact in the context of Lie algebras - this use of compact in Lie algebras is very common, e.g. Barut's Group Representations Ch. 1 defines compactness of a Lie algebra by saying a Lie algebra is compact if there exists a positive-definite quadratic form satisfying $([X,Y],Z) = (X,[Y,Z])$, and the definition comes from (as Barut points out in a footnote) the fact that the Lie algebra of a compact Lie group (in the topological sense) is compact in this sense.
Again, Varadarajan calls it "admits a compact form". This is way better than to abuse an already defined term. But Varadarajan is only a mathematician and not a physicist as O’Raifeartaigh. Yes, it has to do with compact groups which indeed are compact. But the transition of terms falls short in this context.

#### samalkhaiat

Science Advisor
less nonsense. It is wrong and a misinformation!
So, you are from those who think that statements such as “quarks have colours” is “nonsense” and “wrong”.
And please don’t accuse me of make wrong statement because I DO NOT.
I'm not objecting the theorems, I fight against the term compact in this context.
Then take your “fight” to a mathematical court, because the term “compact Lie algebra” was first coined by mathematicians. See S. Helgason’s textbooks and Anthony W. Knapp’s book.
Please stop spreading this false wording! Varadarajan phrases it as "$\mathfrak{g}$ admits a compact form"... It obviously confuses you
How very charming! So the wording “$\mathfrak{g}$ is said to be compact” is wrong because it is not used by Varadarajan, and the wording “$\mathfrak{g}$ is said to be of compact type” is correct because it is used by Varadarajan! Are you, by any chance, related to Varadarajan?

Sir, you do your business with Varadarajan and stop making drama about those who use the terminology of mathematicians other than Varadarajan.

#### fresh_42

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It is misleading and confusing to call $\mathbb{R}^n$ compact. That it s why it is wrong. I told you to name a finite subcover in my example, since it is obviously a three dimensional simple Lie algebra - you may choose whether real or complex - and unless you can do this, it cannot be compact. Period.

Compact is already in use, and of a different meaning. You cannot say, that the very same topological space is compact depending on how you wish to see it. Sorry, but this is more than just sloppy. Even physicists should have limits while messing up mathematical facts.

#### samalkhaiat

Science Advisor
I already said in #17 that "any simple Lie algebra can be put into a compact form if we allow for complex linear transformations (which lead to a different group and algebra). I also showed you how to do that for the simple algebra $\mathfrak{so}(1,2)$. What is the difference between your $\mathfrak{sl}(2, \mathbb{R})$ and my $\mathfrak{so}(1,2)$?
Another mathematical definition of the term compact Lie algebra is "A compact Lie algebra can be regarded as the smallest real form of a corresponding complex Lie algebra".

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