daniele's profile - activity

Hi, I have a problem with ExteriorAlgebra and I do not understand where I am wrong. Let's define an ExteriorAlgebra E with some generators:

E.<e1,e2,e3> = ExteriorAlgebra(SR)

Take for example e1 and compute

e1.interior_product(e1)

which gives 1, correctly. If I instead compute

(e1.interior_product(e1)).constant_coefficient()

this gives 0!

It worked correctly until time ago, this behaviour only appear now: so probably I'm missing something that changed recently. I'm using SageMath version 9.7 on CoCalc.

Thanks in advance!

thanks, corrected!

problem with constant_coefficient of a form of ExteriorAlgebra Hi, I have a problem with ExteriorAlgebra and I do not un

problem with constant_coefficient of a form of ExteriorAlgebra Hi, I have a problem with ExteriorAlgebra and I do not un

Just compute the rank of the matrix:

 matrix([[9,13,5,6,-1],[14,15,-7,-6,4],[-8,-9,12,-5,-9],[-5,-6,-8,9,8],[13,14,15,2,11]]).rank()

Hi,

I have a problem with setting the domains of definition of variables.

For example, typing

var("y") ; assume(y, "real")
conjugate(y + I)

I get the result y + I. The same with var("y", domain="real").

Where I'm wrong?

thanks!

Hi! I come back on an issue on Sage/Maxima, about simplification. I have some large rational polynomial expressions in -say- ten variables that I would like to simplify. simplify() does not work properly: the expression still remains too long. simplify_rational() and simplify_full() are even worse, and too slow. My question is whether, at least for simplifying rational polynomial expression, you know any algorithm or trick that can be used. Thanks in advance!

@paulmasson: thanks, corrected!

Use "reset" command:

reset('f')

Related question: "How can I clear an assignment to a variable?"

Is there a way to compute automorphisms and/or derivations of a Lie algebra (of dimension 4)?

Thanks in advance!

I can not edit, still. Here below a translation:

Write a function in Sage, with the aim to compute the dimension of the vector subspace $W \subseteq \mathrm{End}(M_n(\mathbb{R}))$ constructed as following: W consists of linear maps $F(A;B) \colon M_n(\mathbb{R}) \to M_n(\mathbb{R})$ such that $F(A;B) \colon X \mapsto AX + XB$ with $A$, $B$ in $M_n(\mathbb{R})$. Here, $n\geq 2$.

Hi!

I have a question on how to treat objects depending with parameters.

For example, let M be a matrix depending on a parameter - call it t. The rank command just returns the generic rank. I would like to know the rank of the matrix, varying the parameter. In the example,

_ = var('t')
M = matrix([[t,0],[0,1]])
M.rank()

I would like to get: if t=0, the rank is 1; otherwise is 2. Is there any command for this in Sage? (I've heard about a "full solve" in Mathematica.)

Thanks in advance,

I would like to understand what is the correct and more efficient way for performing sostitution of parameters in objects of some category.

For example, a typical problem for me is the following. I have a 1-form in an ExteriorAlgebra, whose coefficients depends on some variable, let's say t. I find conditions on such parameters, let's say, t should be zero in order that some property holds. (In general, i will have solutions of linear systems.) I try to use "substitute" for setting the parameter to be 0, but this does not work to me..

E.<x,y> = ExteriorAlgebra(SR)
_=var("t")
theta = 3*x+t*y
print "theta = ", theta
theta2 = theta.substitute({t:0})
print "modified theta = ", theta2

Another solution I tried is the following. I assume t==0 at a certain point. But this seems to me to force the substitution before too.

E.<x,y> = ExteriorAlgebra(SR)
_=var("t")
theta = 3*x+t*y
print "theta = ", theta
assume(t==0)
theta2 = theta
print "modified theta = ", theta2

Thanks in advance for any suggestions!

dear Thierry, what kind of assumptions are not taken into account? and what do you mean by "not taken into account"? the problem is related to a reseach problem I'm studying. I should try to write a sample example. Roughly speaking, I have objects in a Lie algebra (whose structure equations may depend on parameters), and I am declaring objects with variable coefficients. Some additional properties add constraints on the possible values of the coefficients. I am using assume() to take into account the several properties. Thanks in advance, all the best,

Hi, how can I recognize an equality from an inequality, in a list of assumptions?

thanks,

Hi,

how can I assume that a variable admits only a finite number of possible values? E.g., the variable x can only be 0 or 1. I could assume that

x*(x-1)==0

but I am looking whether there is a straightforward way, too.

thanks in advance