103 reputation
10
bio website jaivirbaweja.yolasite.com
location Reston, VA
age 17
visits member for 2 years, 6 months
seen Sep 28 '13 at 19:59

I am a young hobbyist mathematician primarily interested in the areas of the study of manifolds such as differential geometry and Kahler geometry. However I am also interested in helping others in mathematics and in expanding my knowledge through that process.


Feb
10
comment Are these vectors in the span of $\mathbb R^3$?
Solving the system to be linearly independent (definition of linear subspace), we get the basis vectors \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}, and \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}, which form the canonical basis (i.e. coordinates are the same) for $\mathbb{R}^3$.
Jan
30
comment On a theorem on Lie derivatives
I follow the first part of your question, but I think that since we have that $i_X: \Omega^{n} \to \Omega^{n-1}$ is definition of interior product, the result follows since $i_X df= f$ and since all zero-forms are trivially zero (need it for the Lie algebra) LHS vanishes.
Jan
27
revised Minimal polynomial over the field $\Bbb Q$
Edited in $\LaTeX$.
Jan
27
suggested suggested edit on Minimal polynomial over the field $\Bbb Q$
Jan
25
awarded  Commentator
Jan
25
comment How can I prove that “If $M$ is contractible differentiable manifold, then $M$ is orientable?”
@user- Please see FAQ page on how to ask a good question.
Jan
25
comment How can I prove that “If $M$ is contractible differentiable manifold, then $M$ is orientable?”
By the Poincare lemma, all closed forms are exact. Then by the natural projection $\pi: TM \to M$.., compute the homology groups and show isomorphic to $\mathbb{Z}.
Jan
25
comment Lim Sup/Inf for real valued functions
The supremum is the least upper bound on $f(x)$. Therefore by definition of limit we have that it is least upper bound on a point. I think that it only exists if $\textup{lim} \ \ textup{sup} f(x)= \textup{lim} \ \textup{sup} f(c)$.
Jan
21
revised Prove using the definition of Lebesgue outer measure
Formatting with $\LaTeX$
Jan
21
suggested suggested edit on Prove using the definition of Lebesgue outer measure
Jan
21
awarded  Autobiographer
Jan
21
revised Sketching complex numbers in coordinate system
Improved $\LaTeX$ formatting so equations easier to understand
Jan
21
suggested suggested edit on Sketching complex numbers in coordinate system
Jan
21
accepted Prerequisites for understanding the Hodge conjecture
Jan
21
comment Curvature of Hyperbolic Space
Given the formula for the sectional curvature, we can make the tangent vectors orthonormal by taking $u= \frac{x^i}{4\delta_{ij}- 8x^i 8x^j}, v= \frac{x^j}{4\delta_{ij}- 8x^i 8x^j}$. Now we can simply use the standard Riemann curvature $K(u,v)= \left \langle R(Du,Dv)Dv,Du \right \rangle$, since the tangent vectors are along the tangent space. This definition still involves the Riemann curvature tensor as required in your question.
Jan
21
revised Specific help in showing that Poisson Bracket is part of this Lie Algebra
Changed the {f,g} to $\LaTeX$
Jan
21
suggested suggested edit on Specific help in showing that Poisson Bracket is part of this Lie Algebra
Jan
21
comment Multilinear Functions
In a multilinear function, each of the seperate variables are linear. For example $f(cx, by, aw)= cf(x)+ bf(y)+af(w)$, where the property that $f(a+b)=f(a)+f(b)$ of a linear map was used along with the other property.
Jan
21
comment How should I prove a set is convex?
Since $S$ is contained in a real vector space, we have to show that for all $t \in [0,1]$, $(1-t)x+ty=1$. The condition $x^Tx=1$ implies that matrix is unitary, and so letting $x_1=x$ and $x_2=y$ in the equation above, you can check that it is convex.
Jan
21
answered Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity?