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Hessian matrix vs differential 2-form
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Could someone clarify the convention that the second derivative of a scalar function $f: Bbb R^n rightarrow Bbb R$ is sometimes defined as a linear operator $D^2f : Bbb R^n rightarrow L(Bbb R^n, L(Bbb R^n, Bbb R))$, which is also identified with the Hessian matrix (filled with second order partial derivatives), and sometimes as a differential 2-form?
It seems that both happens in vector calculus.
Is the definition of derivative just context dependent, and the first example is just a case of "standard derivative" and the second one of the "exterior derivative"?
Or am I really confused about something?
multivariable-calculus vector-analysis differential-forms hessian-matrix
asked Jul 31 at 19:10
Leon Staresinic
61
add a comment |Â
up vote
1
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Could someone clarify the convention that the second derivative of a scalar function $f: Bbb R^n rightarrow Bbb R$ is sometimes defined as a linear operator $D^2f : Bbb R^n rightarrow L(Bbb R^n, L(Bbb R^n, Bbb R))$, which is also identified with the Hessian matrix (filled with second order partial derivatives), and sometimes as a differential 2-form?
It seems that both happens in vector calculus.
Is the definition of derivative just context dependent, and the first example is just a case of "standard derivative" and the second one of the "exterior derivative"?
Or am I really confused about something?
multivariable-calculus vector-analysis differential-forms hessian-matrix
asked Jul 31 at 19:10
Leon Staresinic
61
1
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Could someone clarify the convention that the second derivative of a scalar function $f: Bbb R^n rightarrow Bbb R$ is sometimes defined as a linear operator $D^2f : Bbb R^n rightarrow L(Bbb R^n, L(Bbb R^n, Bbb R))$, which is also identified with the Hessian matrix (filled with second order partial derivatives), and sometimes as a differential 2-form?
It seems that both happens in vector calculus.
Is the definition of derivative just context dependent, and the first example is just a case of "standard derivative" and the second one of the "exterior derivative"?
Or am I really confused about something?
multivariable-calculus vector-analysis differential-forms hessian-matrix
asked Jul 31 at 19:10
Leon Staresinic
61
Could someone clarify the convention that the second derivative of a scalar function $f: Bbb R^n rightarrow Bbb R$ is sometimes defined as a linear operator $D^2f : Bbb R^n rightarrow L(Bbb R^n, L(Bbb R^n, Bbb R))$, which is also identified with the Hessian matrix (filled with second order partial derivatives), and sometimes as a differential 2-form?
It seems that both happens in vector calculus.
Is the definition of derivative just context dependent, and the first example is just a case of "standard derivative" and the second one of the "exterior derivative"?
Or am I really confused about something?
multivariable-calculus vector-analysis differential-forms hessian-matrix
asked Jul 31 at 19:10
Leon Staresinic
61
asked Jul 31 at 19:10
Leon Staresinic
61
asked Jul 31 at 19:10
Leon Staresinic
61
asked Jul 31 at 19:10
Leon Staresinic
61
61
1
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36
add a comment |Â
1
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36
1
1
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36
add a comment |Â
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1
The second derivative is definitely not a differential $2$-form. Applying the exterior derivative twice gives zero.
– Qiaochu Yuan
Jul 31 at 21:11
Not necessarily, if the form isn't $C^2$. So, the difference is that it's simply a completely different operation?
– Leon Staresinic
Aug 1 at 8:36