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I have an account here mostly so that I can ask stupid questions. Hopefully over time my questions will become less stupid, or at least more complicated and therefore laughably naive only to a select few experts.

I am an undergraduate. Most or all of my questions here are from self-study, so I really appreciate the answers given.


Sep
12
comment On the Lebesgue measure of the set of small values of an analytic function on C
No, consider $1 / z$. Then $m(U_{\epsilon})$ is constantly infinity.
Sep
1
accepted Commutative lie groups - how is $(R, >, 1)$ a $T^q \times R^p$
Sep
1
asked Why is the Leibniz rule a sufficient ingredient in the construction of the tangent space?
Sep
1
asked What is the use of scheme theory?
Aug
31
comment What is the right category in which to think of adjoints?
So the data of an adjunction is a natural transformation between the functors $Mor(F(A),B)$ and $Mor(A,G(B))$ from $C^{op} \times D \to Set$ (given that $F$ and $G$ are functors)?
Aug
31
accepted What is the right category in which to think of adjoints?
Aug
31
accepted What does $GL_n(R)$ look like?
Aug
31
asked What is the right category in which to think of adjoints?
Aug
20
comment What do the right and left translation maps on $GL_n(R)$ look like?
@MichaelAlbanese Okay, point taken.
Aug
20
comment What do the right and left translation maps on $GL_n(R)$ look like?
I don't see a good way to choose between the two answers. They are both very different, and very good. Neither is "correct," because my question didn't really admit a clear answer.
Aug
20
comment What do the right and left translation maps on $GL_n(R)$ look like?
@MichaelAlbanese Do people actually care about these points?
Aug
20
asked What do the right and left translation maps on $GL_n(R)$ look like?
Aug
19
comment What does “Prove by induction” mean?
"Mathematics is only an apparatus, not a source of knowledge itself." I think this is kind of extreme. I guess it depends on what you consider to be knowledge.
Aug
19
comment What does $GL_n(R)$ look like?
@QiaochuYuan Oh - no, you're completely right. You just choose one splitting. Of which there are many. For example, fixing a positive branch of the nth root of the absolute value, and using some fixed diagonal entry to account for sign.
Aug
19
comment What does $GL_n(R)$ look like?
@QiaochuYuan Thanks. You're right - I don't think that I can just naively take the nth root either. I'll work this out as an exercise then. That would probably be good for me.
Aug
19
comment What does “Prove by induction” mean?
I'm not sure what you mean by your second sentence.
Aug
19
comment What does $GL_n(R)$ look like?
Presumably because a matrix in $GL(n)$ can be decomposed as $\lambda B$, where $\lambda$ is its determinant and $B \in SL(n)$? The determinant (representing the coset of $GL(n) / SL(n)$) and the representative in $SL(n)$ are well defined and smooth projection maps. (The later is smooth because the determinant is smooth and nonzero, so we can invert it to get the matrix $\lambda^{-1} I$. And since $GL(n)$ is a Lie group, multiplication by this matrix is smooth.) I think. So now one could argue that $GL(n)$ (with these projections) satisfies the universal property for products.
Aug
19
comment What does “Prove by induction” mean?
"Mathematical induction adds nothing new to human knowledge." I think this is somewhat biased, and also probably wrong. I can think of a number of theorems that are very easily shown through induction, but which are probably cumbersome to express otherwise. (Ex. All the basic theorems about p-groups, for instance.)
Aug
19
comment What does $GL_n(R)$ look like?
What can one do to understand the geometry of $SL_n(R)$? Its complement is no longer a variety, at least in a way that is obvious to me.
Aug
19
comment What does $GL_n(R)$ look like?
Thanks. So the fact that the complement is in general a conical hypersurface follows from the determinant being a homogenous polynomial of degree n?