Marek
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 Jan22 answered Determine whether the given function is an integrating factor for the DE Jan22 comment Adjoint operator on subspace That's not much better for two reasons. First, $f$ can belong to $H_2 \setminus V_2$, so the expression is again not defined and second, $T$ need not be either injective or surjective, so $T^{-1}$ is not defined in general and even when it is, not for all $f \in V_2$. I think not much can be said about your question unless you add more requirements either on some of the spaces or on the map $T$. Jan22 comment What is the last index of a third-order tensor called? @Giuseppe: I agree with your sentiment of giving names but I stand by what I said. If the question were reformulated so as to use the word matrix instead of tensor (since it moreover doesn't really involve tensors at all -- it's only a question on the terminology of generalized matrices), I will remove the downvote. But I'll leave it here precisely because I am not a fan of the engineering attitude of thinking about the tensor as a matrix (of course, you can treat tensors that way, but then, you can treat almost anything that way, so that's not exactly a great argument..). Jan22 comment Adjoint operator on subspace What do you mean interpret? Also, how could it possibly be $Tf$ when $f \in V_2^*$ while $T$ acts on the elements from $V_1$? This expression would not make any sense. Is there something you are not telling us? Jan22 answered Another simple/conceptual limit question Jan22 comment What is the last index of a third-order tensor called? -1 Tensor is not a matrix, so the question doesn't really make sense. Jan22 answered Connection between Euler characteristic and degree of the Gauss map Jan22 answered Is there some difference between the two definition of integration along curves about complex function Jan19 comment The solution of the following two equations in two unknowns Is this a homework? Also, what have you tried and what kind of help (e.g. full solution) do you expect? Be more specific.. Jan19 comment topological group operation vs homotopy group operation Ah, I see, I misunderstood your construction. Sorry. Jan18 comment Invariants of representation theory of Lie groups Special linear group has determinant equal to one by definition, so I have no idea what you are asking. For the second part, if by generator you mean an element of the representation of the corresponding Lie algebra (the derived representation) then this vanishes since $\exp {\rm tr} A = \det \exp A$. Jan18 comment topological group operation vs homotopy group operation I don't see how to use E-H here since you don't have two operations on one set but here each operation is on a different set. Anyway, interesting question and I'm curious about the answer in either direction. Can we say anything concrete e.g. about $SU(2)$? Jan18 revised Computing derivative $x^{x^x}$ Fixed an incorrect formula Jan18 suggested approved edit on Computing derivative $x^{x^x}$ Jan18 comment Monic polynomial divided by $x-r$ I didn't notice that assumption, sorry. Nevertheless, I think it's worth pointing out that the assumption that $R$ be a field is not necessary here and that recalling general division algorithm is quite an overkill when dividing by linear monic. Jan18 comment Monic polynomial divided by $x-r$ Your $(1)$ is not true in an arbitrary ring. For example consider $x^2 / 2x$ in $(\mathbb Z / 4 \mathbb Z)[x]$. It holds however when dividing by the monic polynomials, which is the case here. Jan18 answered Monic polynomial divided by $x-r$ Jan18 revised Multiple Choice question about a continuous function Added further discussion Jan18 answered Multiple Choice question about a continuous function Jan18 comment ${\mathbb Q}$ as a direct product. This seems like a very strange idea. To prove that $\mathbb Q$ is not a direct product, you must first and foremost use properties of $\mathbb Q$ itself. Can you point out where you are doing this? This kind of argument is necessary because there can't be any uniform categorical argument for general group since there are many groups which are direct products and many which aren't. On the other hand, I don't see any advantage in translating the properties of $\mathbb Q$ (e.g. characterizing it by some universal property) to categorical language.