Jaivir Baweja

103 reputation

bio	website	jaivirbaweja.yolasite.com
	location	Reston, VA
	age	16
visits	member for	1 year, 1 month
	seen	2 days ago
stats	profile views	57

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.

	103 reputation	bio	website	jaivirbaweja.yolasite.com	visits	member for	1 year, 1 month
10 badges		location	Reston, VA		seen	2 days ago

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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?