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2

I just want to advertise this counterexample due to Robert Bonic, which I find much more appealing than Kupka's. Consider $E=C^0([0,1])$ and $G:E\rightarrow E$ which sends $G(f)(x)=f(x)^3$. Then $G$ is smooth and $dG(f)(u)=3f^2 u$. Thus if $f$ has a zero, $f$ is a critical point for $G$. Now set $$ C=\{f\in E\,|f(x)<0,\quad f(y)>0,\quad \text{for ...


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Expanding by linearity and removing all terms we know are zero, and using symmetry to gather the rest of the terms, we get $$\begin{align} 0 = &\,\frac13T(\vec v_1 + \vec v_2 + \vec v_3,\vec v_1 + \vec v_2 + \vec v_3, \vec v_1 + \vec v_2 + \vec v_3)\\ =&\, T(\vec v_1, \vec v_1, \vec v_2)+ T(\vec v_1, \vec v_2, \vec v_2) \\ &+ T(\vec v_1, \vec ...


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I found the definition to $k$-linear symmetric, here Representation of symmetric functions. And the norm $||.||_{k}$, http://www.math.uni-oldenburg.de/preprints/get/source/flo-gar.pdf page 2. Just change $k=2$ for $k$ arbitrary.


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First, it's pretty clear that the field $K$ does not matter in any way (if you know how to prove things for $\mathbb{R}$, then just check that you don't use any special property of this field). Then, be careful : the statement for tensor algebras is already false for finite-dimensional vector spaces : if $V$ has finite dimension, $T(V^*)$ has countable ...



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