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	<title>Notes for AKT-140224/0:27:53 - Revision history</title>
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	<updated>2026-05-07T17:06:10Z</updated>
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		<id>https://drorbn.net/index.php?title=Notes_for_AKT-140224/0:27:53&amp;diff=16593&amp;oldid=prev</id>
		<title>Leo algknt: Created page with &quot;&#039;&#039;&#039;Lie algebra of dimensions 1 and 2&#039;&#039;&#039;  1.  &#039;&#039;&#039;one-dimensional Lie algebras&#039;&#039;&#039;  are unique up to isomorphism. For if &lt;math&gt;\mathfrak{g} = \langle x \rangle &lt;/math&gt; is a one d...&quot;</title>
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		<updated>2018-07-04T06:27:24Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;#039;&amp;#039;&amp;#039;Lie algebra of dimensions 1 and 2&amp;#039;&amp;#039;&amp;#039;  1.  &amp;#039;&amp;#039;&amp;#039;one-dimensional Lie algebras&amp;#039;&amp;#039;&amp;#039;  are unique up to isomorphism. For if &amp;lt;math&amp;gt;\mathfrak{g} = \langle x \rangle &amp;lt;/math&amp;gt; is a one d...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;#039;&amp;#039;&amp;#039;Lie algebra of dimensions 1 and 2&amp;#039;&amp;#039;&amp;#039;&lt;br /&gt;
&lt;br /&gt;
1.  &amp;#039;&amp;#039;&amp;#039;one-dimensional Lie algebras&amp;#039;&amp;#039;&amp;#039;  are unique up to isomorphism. For if &amp;lt;math&amp;gt;\mathfrak{g} = \langle x \rangle &amp;lt;/math&amp;gt; is a one dimensional Lie algebra, then since the bracket is antisymmetric, we have &amp;lt;math&amp;gt;[x, x] = 0&amp;lt;/math&amp;gt;. Thus the bracket is zero and &amp;lt;math&amp;gt;\mathfrak{g}&amp;lt;/math&amp;gt; is unique up to isomorphism.&lt;br /&gt;
&lt;br /&gt;
2. &amp;lt;math&amp;gt;\mathfrak{g} = \mathbb{F}^2 = \{ax + by \;|\; a, b \in  \mathbb{F}\}&amp;lt;/math&amp;gt;. &amp;lt;math&amp;gt;\mathfrak{g}&amp;lt;/math&amp;gt; is a two-dimensional Lie algebra. There are only two of such up to isomorphism, that is, the one with the bracket equal to zero and the other with bracket &amp;lt;math&amp;gt;[x, y] = x&amp;lt;/math&amp;gt;.&lt;/div&gt;</summary>
		<author><name>Leo algknt</name></author>
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