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Edit: According to the comment by @LSpice we realise the existing link to the main motivation of the question is not available. Then we search for the paper we found the following version:

https://www.semanticscholar.org/paper/Models-of-Consumer-Behaviour-Problem-presented-by-Schlijper/6e4edeaad6e400b4acee7a2ed180fb73dfa1f18e

And this link too:

https://www.researchgate.net/publication/277825529_Models_of_consumer_behaviour#fullTextFileContent


We define $$L_{n}=\{A=(a_{ij})\in M_{n}(\mathbb{R})\mid \sum_{i=1}^{n} a_{ij}=0 \;\;\;\text{for every fixed j}\}$$ This is a Lie subalgebra of $M_{n}(\mathbb{R})$.

A dynamic-geometric proof for this fact is that;

$"$ The affine hyper plane containing the standard simplex $\Delta^{n-1} \subset \mathbb{R}^{n}$ is invariant under the flow of a linear vector field $\dot X =AX,\;\; A\in M_{n}(\mathbb{R})$ if and only if $A\in L_{n}"$. On the other hand the space of all (linear) vector fields tangent to a submanifold (say the affine hyper plane containing $\Delta ^{n-1}$) of any manifold (say $\mathbb{R}^{n}$) is closed under Lie braket.

Is there any name-notation for this particular lie algebra? What is a precise Lie group whose Lie algebra is isomorphic to $L_{n}$? What is the precise description of the image of $L_{n}$ under the exponential map $A \mapsto e^{A}$?

What are some obstructions for embedding of a finite dimensional Lie algebra in some $L_{n}$?

Edited at Nov. 28, 2017, Motivation for consideration of such an $L_n$:

More than two years ago, an economist very kindlyintroduced me this paper Model of consumer behaviour and suggests me to generalize the mathematical part of the paper. My observation was the following fact as generalization of the above paper:

Fact: The linear equation $X'=AX$ has a unique equilibrium on the standard simplex $\Delta^n$ if $A=(a_{ij})$ belongs to $L_n$ and $a_{ij}>0\; \text{for}\; i\neq j$. Moreover the equilibrium is an attractor singularity.

Note: In the linked paper, this fact is proved for $n=2,3$ with quite computational method.

Sketch of Proof for arbitrary $n$: The system is a monoton dynamical system since the off diagonal elements of the jacobian of the vector field is positive.

This implies that $\Delta^{n-1}$ is flow invariant. So the flow version of the Leray Schauder fiexed point theorem implies that the standard simplex has at least one singularity. Now this post implies that the singularity is unique since the matrix $A$ is a matrix of rank $n-1$. To prove that the singularity is attractor we use again the monotone property of the flow as follows.

Note that the flow $\phi_t$ of a monotone dynamical system satisfies $x \leq y \implies \phi_t(x) \leq \phi_t(y)$. The order we are considering is defined as follows:

$$x\leq y \iff x_i \leq y_i,\;\; \forall i$$ where $x_i, y_i,s$ are coordinates of $x,y$, respectively.

Now we prove that the unique singularity of $X'=AX$ in the standard simplex is an attractor singularity provided that $A \in L_n$ and its off diagonal entries are strictly positive.

Let $\ell$ be the one dimensional subspace which determines the kernel of $A$. Then $\ell$ intersect the standard simplex at point $p\in \Delta^{n-1}$. The point $p$ is the unique singularity of the system on the simplex. Lets choose a point $q \in \ell$ where $q$ is near to $p$ and $\sum q_i<1$. We construct a cone with base point $q$. By monotone property of flow we conclude that this cone is flow invariant. If we let $q$ approach to $p$, then we observe that $p$ is an stable singularity. By a simple argument one can shows that the singularity can be approached by a band of closed orbit because the direction field on the boundary of the simplex is toward interior. So the unique singularity is attractor.

The " Monoton dynamical systems" is introduced by Morris W. Hirsch.

Here is a picture for our proof:

enter image description here

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    $\begingroup$ Your link to "this paper" is now broken. $\endgroup$
    – LSpice
    Commented Jul 9, 2020 at 13:37
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    $\begingroup$ @LSpice yes I see. The title of the paper is "A model of consumer bahavior" I try to find another link containing this paper. On this paper they showed rhat for n=2 or 3 any such $A\in L_n$ has a unique equilibrum on $\Delta^n$. They used the direct computational method. But the interesting gact is that for every $n$ we have a unique atractive equilibrum in $\Delta^n$. The key point which was not observed by the authors of the linked paper is that the system satisfies the Hirsch competitive dynamical property(monoton dynamical system"). It is a little bizzare that yhe.linked paper doesnit $\endgroup$ Commented Jul 11, 2020 at 9:40
  • $\begingroup$ does not work(it is broken) $\endgroup$ Commented Jul 11, 2020 at 9:41
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    $\begingroup$ Is "A model of consumer behavior" the full title? Who are the authors? $\endgroup$
    – LSpice
    Commented Jul 11, 2020 at 13:08
  • $\begingroup$ @LSpice Yes I just find i add to the revised version of question. Thank you for your attention $\endgroup$ Commented Jul 11, 2020 at 13:12

1 Answer 1

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Your Lie algebra consists of $X$ such that $Xv = 0$ where $v$ is the all-ones vector. So the corresponding Lie group in $GL_n(\mathbb{R})$ consists of $g \in GL_n(\mathbb{R})$ such that $gv = v$.

This is conjugate to the same group but where $v$ has been replaced with the vector $e_1 = (1, 0, 0, \dots) \in \mathbb{R}^n$, which is easier to deal with. The corresponding subgroup of $GL_n(\mathbb{R})$ consists of invertible matrices whose first column is $1, 0, 0, \dots$. This is the general affine group. Abstractly it is the semidirect product $\mathbb{R}^{n-1} \rtimes GL_{n-1}(\mathbb{R})$ where the first factor comes from the first row and the second factor comes from the rest. By Ado's theorem every Lie algebra embeds into some $\mathfrak{gl}_{n-1}(\mathbb{R})$ and hence into the corresponding semidirect product.

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  • $\begingroup$ Thank you very much for your answer. I think you mean $v^{tr}X=0$ ? So a related question: is every Lie subalgebra $L$ of $M_{n}(\mathbb{R})$ isomorphic to its transpose $L^{tr}$ while the transpose operator is not Lie algebra automorphism? $\endgroup$ Commented Dec 11, 2015 at 6:40
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    $\begingroup$ @Ali: $X \mapsto - X^T$ is a Lie algebra automorphism of $\mathfrak{gl}_n$, which is the Lie algebra version of $g \mapsto (g^T)^{-1}$ being a Lie group automorphism of $GL_n$. $\endgroup$ Commented Dec 11, 2015 at 6:42
  • $\begingroup$ motivated by your answer, can one consider $\{ (v,g)\in S^{n-1}\times Gl(n)\mid g.v=v \}$, as total space of a natural fiber or principal fiber bundle over $S^{n-1}$? $\endgroup$ Commented Dec 11, 2015 at 7:40
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    $\begingroup$ In $GL_n$ the corresponding Lie group is $\mathbf{R}^{n-1}\rtimes GL_{n-1}^+(\mathbf{R})$ (which has index 2 in the non-connected $\mathbf{R}^{n-1}\rtimes GL_{n-1}(\mathbf{R})$). This group has a trivial center; its $\pi_1$ is the same as in the maximal compact subgroup $S0(n-1)$ which has $\pi_1$ of order $1$ if $n=2$, cyclic infinite if $n=3$, and of order 2 if $n\ge 4$. This allows to list those connected Lie groups having $L_n$ as Lie algebra. $\endgroup$
    – YCor
    Commented Dec 11, 2015 at 13:19
  • $\begingroup$ @YCor Thank you for your interesting comment which completes the answer. What about my other question in the previous comment? $\endgroup$ Commented Dec 12, 2015 at 7:38

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