# Lie theory - sum of nilpotent ideals is nilpotent?

• Silversonic
In summary, the proof shows that the sum of two nilpotent ideals is nilpotent, but the step of induction gets stuck for one person.
Silversonic
I can't wrap my head around this proof that the sum of two nilpotent ideals is nilpotent, I get stuck at one stage:

http://imageshack.com/a/img706/5732/5wgq.png

I'm fine with every except showing by induction $(I+J)^{N+k} = I^k \cap J + I \cap J^k$. Here's my attempt;

Base case: k = 1,

$(I+J)^{N+1} = [I+J, (I+J)^N] \subseteq [I+J,I \cap J] = [I, I \cap J] + [J, I \cap J] \subseteq I \cap J + I \cap J$

since $[I, I \cap J], [J, I \cap J] \subseteq I \cap J$ as $I \cap J$ is an ideal.

Now inductive step;

$(I+J)^{N+k+1} = [I+J, (I+J)^{N+k}] = [I+J, I^k \cap J + I \cap J^k] = [I, I^k \cap J] + [J, I^k \cap J] + [I,I \cap J^k] + [J,I \cap J^k]$

Now it's easy to see

$[I, I^k \cap J] \subseteq I^{k+1} \cap J$

$[J, I \cap J^k] \subseteq I \cap J^{k+1}$

But I have no idea what I can do with the $[J, I^k \cap J] + [I,I \cap J^k]$ term so that it reduces to the form I want. Any help?

Last edited by a moderator:
Are you stuck using this exact argument? Unless I have misunderstood something the proof should go through easily with a slight modification: Essentially after repeatedly multiplying I∩J by I+J we should end up with something like ∑Ij∩Jk-j and choosing k so large that either Ij or Jk-j vanishes for each choice of j should complete the argument. Maybe I have misunderstood something here though and this fails for some reason.

jgens said:
Are you stuck using this exact argument? Unless I have misunderstood something the proof should go through easily with a slight modification: Essentially after repeatedly multiplying I∩J by I+J we should end up with something like ∑Ij∩Jk-j and choosing k so large that either Ij or Jk-j vanishes for each choice of j should complete the argument. Maybe I have misunderstood something here though and this fails for some reason.

You're quite right, and that's a different way of looking at it than I did. I think by this method it shows it is zero when $k = 2N$ while the one in my screenshot shows it for lower, at $k = N$. It doesn't matter though since my goal is to show it's just nilpotent, thanks.

## 1. What is Lie theory?

Lie theory is a branch of mathematics that studies continuous symmetry and transformation groups. It is named after the mathematician Sophus Lie, who laid the foundations for this theory in the 19th century.

## 2. What are nilpotent ideals?

Nilpotent ideals are subspaces of a Lie algebra that satisfy a certain property called "nilpotency". This means that repeatedly applying the Lie bracket operation (which measures the failure of a vector field to commute with another vector field) to elements of the ideal eventually results in zero.

## 3. What does it mean for a sum of nilpotent ideals to be nilpotent?

If the sum of two or more nilpotent ideals is also a nilpotent ideal, it means that repeatedly applying the Lie bracket operation to elements of the sum eventually results in zero. In other words, the sum of nilpotent ideals is itself a nilpotent ideal.

## 4. Why is the sum of nilpotent ideals important in Lie theory?

The sum of nilpotent ideals is important in Lie theory because it allows for the study of Lie algebras and their structure. By understanding the properties of nilpotent ideals and their sums, we can gain insight into the structure and properties of Lie algebras in general.

## 5. How is the property of nilpotency useful in Lie theory?

The property of nilpotency allows for the classification and study of different types of Lie algebras. It also helps in understanding the structure of Lie algebras, their representations, and their applications in physics, engineering, and other areas of mathematics.

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