Why is the gradient a covector?

Vectors in geometry

Why is the gradient a covector?

Postby Guest » Sun Dec 30, 2018 11:20 pm

Why is the gradient a covector?

I understand that a covector is a map from a vector to a real number.
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Re: Why is the gradient a covector?

Postby Guest » Sun Dec 30, 2018 11:21 pm

how do I get a respnse here? my email is drlang@tx3.net

Thanks
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Re: Why is the gradient a covector?

Postby Guest » Fri Jun 11, 2021 10:23 am

People aren't likely to give you a personal answer by e-mail. Post an intelligible question here and you will get an answer here. You ask "Why is the gradient a covector". The problem is that the gradient isn't a covector! The "divergence" is a co-vector.

You say "I understand that a covector is a map from a vector to a real number." Okay, do you understand what a "gradient" is? in three dimensions the gradient if a differential operator: [tex]\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}[/tex]. That maps a a scalar function to a vector.

The divergence of a vector is [tex]\nabla\cdot \vec{v}= \frac{\partial v_x}{\partial x}+ \frac{\partial v_y}{\partial y}+ \frac{\partial v_z}{\partial z}[/tex]. That maps a vector to a scalar so, by your definition, is a "covector" (Strictly speaking the scalar does not have to be a "real number". It could be a complex number.)

Just to be complete, there is also the "curl", [tex]\nabla\times \vec{v}= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_x & v_y & v_z \end{array}\right|[/tex] which maps a vector to a vector.
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