I've been learning some introductory analysis on manifolds and have had a small issue ever since the notion of tangent spaces at points on a differentiable manifold was introduced.

In our lectures, we began with the definition using equivalence classes of curves. But it is also possible to define tangent spaces using derivations of smooth functions (and apparently several other ways too, but for now I'm only familiar with these two).

It seems intuitively sensible to call both these pictures (the curve and derivative ones) "equivalent": let the point of interest be $p$ and pick a local chart $\phi$. Then we form a quotient of the set of curves through $p$ (parametrized so that $p=\gamma(0)$), declaring $\gamma_1\sim\gamma_2$ iff $(\phi\,\circ\,\gamma_1)'(0)=(\phi\,\circ\,\gamma_2)'(0)$. This is one particular version of a tangent space at $p$. But we could also define it as the space of derivations, i.e. linear maps from $C^\infty(M)$ to $\mathbb{R}$ satisfying the Leibnitz rule $$D(fg)=D(f)g(p)+f(p)D(g)$$ For any equivalence class of curves $[\gamma]$ at $p$, the operator defined on $C^\infty(M)$ by $$ D_{[\gamma]}(f)=(f\circ\gamma)'(0) $$ is a derivation; conversely, it is true that every derivation is such a directional derivative (proof: Equivalence of definitions of tangent space).

Most of this a recap of a part of Wikipedia. At any rate, both of these notions seem to give in some sense "the same" tangent spaces.

Here is my problem: I don't actually understand what precisely it is we are checking for when trying to decide if some two definitions are equivalent; right now, all I would personally try to do is show isomorphism of vector spaces and then try to convince myself that this isomorphism respects some vague notion of direction. But then $\mathbb{R}^{\mathrm{dim}(M)}$ is certainly isomorphic to any tangent space of the manifold $M$, at least as a vector space. Nevertheless, just declaring $T_pM=\mathbb{R}^{\mathrm{dim}(M)}$ doesn't strike me as a successful construction of a tangent space.

Now, there are two levels to my question, ordered by "degree of abstraction", so to speak (presumably they also get harder to answer). I do, however, believe they are connected.

First, is there some precise notion of vector space isomorphisms respecting direction on a manifold? Specifically, is $\mathbb{R}^{\mathrm{dim}(M)}$ a valid tangent space or is it not, or do I perhaps have to specify some additional structure on it and then check that the additional structure relates to, say, the curve definition in a correct way? (I suppose this last case would require taking one definition of the tangent space as the absolute foundation and comparing all others to it, which I find somewhat unsatisfying.)

Second, is there perhaps an abstract, "external" definition of a tangent space? What I'm talking about could be something like, "Given a smooth manifold $M$, a point $p\in M$ and a vector space $V$, this vector space is called a tangent space at $p$ if it satisfies some properties $X,Y,Z...$" where these $X,Y,Z$ don't depend on the type of objects in $V$ or other particular details specific to $V$.

The motivation behind asking this is related to the situation with ordered pairs of objects (yes, this is quite a leap): I can use the Kuratowski definition or infinitely many others, and in each case, I will be able to eventually convince myself that, indeed, this thing before me works just as well to encode "ordered-ness" of objects as any other. But I don't have to keep referring to one of these specific cases, I just need to describe how pairs should arise and behave in general: there is a two-place function $f$ that sends two objects $x$ and $y$ to $(x,y)$ and there are two projections $\pi_1,\pi_2$ that pull $x$ and $y$ back out. (For a precise definition see this PDF, I summarised the discussion from there. It goes on to define products also within category theory.) Furthermore, I would find it highly suspect if some theorem about ordered pairs referred to the particulars of the Kuratowski definition - all the relevant information about $(x,y)$ should be recoverable from just the abstract setup described above (or better yet, in the linked PDF). Is there some way of treating tangent spaces in this same spirit?

I know this question is vague, but I honestly don't know how better to phrase it, I hope I've at least gotten the mindset across if nothing else.


3 Answers 3


1) The tangent space $T_p M$ is a vector space and, as you pointed out, any two vector spaces of the same dimension are isomorphic. There are two issues that make the definitions in the post nontrivial. First, that isomorphism is not canonical; it depends on (e.g.) a choice of basis. Second, and more importantly, the right notion here is that of the tangent bundle versus the tangent space. That is, the tangent bundle $TM$ is a space together with a (continuous) projection $\pi:TM \to M$ such that every point $p\in M$ has a neighborhood $U$ above which $\pi$ is just the projection $U \times \mathbb{R}^n \to U$ for some fixed $n$. In this case, $TM$ is the collection of the $T_p M$ for $p\in M$, topologized in a certain way.

2) There's nothing inherently wrong with having a lot of equivalent definitions of the tangent space; consider all the different definitions of an ordinary derivative. Ultimately, all of these definitions come down to the fact that a tangent space is defined locally (i.e., $T_p M$ only depends on a neighborhood of $p$), and points on manifolds have neighborhoods homeomorphic (in whatever category we're considering, and presumably at least $C^1$ here) to $\mathbb{R}^n$. On $\mathbb{R}^n$, the idea of a tangent space is simple: it's just $\mathbb{R}^n$ itself. The different definitions are just ways of turning that idea into something that doesn't depend on explicit choices of local charts. For motivation, you might want to consider the case where $M$ is smoothly embedded in some $\mathbb{R}^n$. (By the Whitney embedding theorem, this is a trivial assumption, at least if we're assuming second-countability. The trick is coming up with a definition that's independent of that embedding.)

3) As for an abstract or external definition, define the cotangent space $T_p^* M$ to be the quotient $I/I^2$, where $I$ is the space of smooth maps $f:M \to \mathbb{R}$ that vanish at $p$. (It would probably be cleaner to work with the sheaf of smooth functions defined on a neighborhood of $p$, but we can reduce to the case above via a suitable bump function.) The tangent is then the dual of $T_p^* M$, but $T_p^* M$ itself is useful in, for example, defining differential forms.

Beyond that, it sounds like the abstraction you may be looking for (though, unfortunately, it's not particularly category-theoretic) is that of a vector bundle or, more abstractly, a general fiber bundle. The full definition is on (e.g.) wikipedia, but the idea is the same as the one in part (1) above: A bundle with fiber $F$ over a manifold $M$ is a space $E$ along with a continuous surjection $\pi:E \to M$ that locally looks like the projection $U \times F \to U$ onto the first coordinate. The Moebius strip, for example, is a $[0, 1]$-bundle over the circle: It just looks like $[0, 1] \times U$ around a small neighborhood $U$ of a point in the central circle, but the whole space isn't just $[0, 1]\times S^1$.

This turns out to be an extraordinarly useful idea, and it leads to extremely productive ideas about exact sequences in algebraic topology, characteristic classes, classifying spaces, and so on.

  • 4
    $\begingroup$ This is a very nice answer. I do have some quibbles: 1.,2.) I get that the isomorphism is artificial-looking in that an arbitrary choice of basis has to be made, but my confusion comes from the other direction: why is the isomophism between curves and derivations considered canonical when the $\mathbb{R}^{\mathrm{dim}(M)}$ one isn't? I could always choose to put a minus sign here: $-(\phi\circ\gamma)'(0)$ and get something ostensibly just as good. 3.) This is a nice, clean approach. Still, fundamentally it seems to me that we've just come up with another concrete object, something... $\endgroup$
    – J_P
    Commented Mar 13, 2020 at 12:18
  • 1
    $\begingroup$ ...construced in some particular, specific way. (Also, I thought the dual isomorphism was basis-dependent? Though there does seem to be a "normal" way of doing it here: mathoverflow.net/questions/88880/…) I suspect these issues just come down to the fact that I don't understand what "canonical isomorphism" means in this context, strictly formally speaking. At any rate, this answer has given me a lot to chew on. I'll wait a while to see if anyone else chips in, otherwise I'll accept this. $\endgroup$
    – J_P
    Commented Mar 13, 2020 at 12:19
  • 3
    $\begingroup$ For (1) and (2), maybe "canonical" isn't the right word there. Just setting $T_p M = \mathbb{R}^n$ tells us nothing about $M$ besides its dimension; it would be the same object (not just up to isomorphism, but literally the same object) for any manifold. What we actually care about is how the $T_p M$ change continuously as a function of $M$; that is, what the tangent bundle structure looks like. The curve definition and others give us there. As for changing the sign of $(\phi \circ \gamma)'(0)$, you'd still get the same space, just a different map into it. $\endgroup$
    – anomaly
    Commented Mar 13, 2020 at 13:01
  • 1
    $\begingroup$ For the issue with the dual basis, the difference is that we can recover $T_p M$ from $T_p^* M$ by looking at maps $T_p^* M \to \mathbb{R}$ (i.e., the double dual) on the tangent bundle rather than arbitrarily on each tangent space. That is, we can choose an isomorphism that varies continually with $p$, which is really all we need. Besides, the cotangent space or bundle is often the object we want in the first place, as with smooth differential forms. $\endgroup$
    – anomaly
    Commented Mar 13, 2020 at 13:10
  • 1
    $\begingroup$ By the way, another motivation for the tangent space definition you might find useful is the constant rank theorem: If $f:M\to N$ is a smooth map between manifolds such that the derivative $Df$ has constant rank $k$ near $p\in M$, then $f$ is of the form $f(x_1, \dots, x_n) = (x_1, \dots, x_k, 0, \dots, 0)$ near $p$ for suitable coordinate charts around $p$ and $f(p)$. Translating that into the case of a nice smooth embedding $M\to \mathbb{R}^m$ should make the curve definition behave according to intuition. $\endgroup$
    – anomaly
    Commented Mar 13, 2020 at 13:17

I think what OP wants is a definition using the concept of functors. Below is the repost of my answer https://mathoverflow.net/q/429683 to the question "An easy way to to explain the equivalence definitions of tangent spaces?"

In short: To give a definition of tangent spaces, one actually define a functor from the category of (pointed)manifolds to the category of vector spaces that is a natural extension of the tangent space of Euclidean spaces satisfying the principle of localization(see below). It can be shown that the tangent space functor is uniquely determined by these two conditions, up to a natural isomorphism.

Hence to check that a concrete definition of tangent space is reasonable, one only need to show that this definition generalizes the concept of tangent spaces of Euclidean spaces, and satisfies the principle of localization. If you have two such definitions, then they are equivalent in the sense that there exists a natural isomorphism between the functors representing the two definitions.

In detail: The category of pointed manifold $\mathcal M$ has objects in the form $(x,M),$ where $M$ is any smooth manifold and $x$ an element of $M.$ A morphism from $(x,M)$ to $(y,N)$ is an ordered quadruple $(x,f,M,N),$ in which $f$ is a functional relation on $M\times N$ with $\mathrm{dom}(f)$ a (not necessarily open)neighborhood of $x,$ $f$ differentiable at $x$ and $f(x)=y.$

More exactly, $f$ is a subset of $M\times N$ such that if $(x,y),(x,y')\in f$ then $y=y'.$ By definition, if $O$ is an open neighborhood of $x$ in $M$ such that $\mathrm{dom}(f)\subset O,$ and $O'$ is an open subset of $N$ with $\mathrm{Im}(f)\subset O',$ then $(x,f,O,O')$ is a morphism from $(x,O)$ to $(y,O').$

The composition of morphisms is given by \begin{equation} \boxed{(y,g,N,Q)\circ (x,f,M,N):=(x,g\circ f,M,Q)} \end{equation} where $g\circ f:=\{(x_1,x_3)|\ \exists \ x_2\in N,\ s.t. \ (x_1,x_2)\in f,\ (x_2,x_3)\in g \}$ is the composition of relations.

If we consider Euclidean spaces only, then we get a full subcategory $\mathcal M(\mathbb E)$ of $\mathcal M.$ Now we define the canonical tangent space of $(x,\mathbb R^n)$ to be simply $\mathbb R^n,$ and the canonical tangent map $T_{\rm canonical}(x,f,\mathbb R^n,\mathbb R^m):=D_xf,$ then we get a functor $T_{\rm canonical}:\mathcal M(\mathbb E)\to \mathsf{Vec}(\mathbb R).$ \begin{equation} \boxed{T_{\rm canonical}\big[(x,\mathbb R^n)\stackrel{(x,f,\mathbb R^n,\mathbb R^m)}{\longrightarrow} (y,\mathbb R^m)\big]:=\mathbb R^n \stackrel{D_xf}{\longrightarrow}\mathbb R^m} \end{equation}

A tangent space functor defined on the full category $\mathcal M$ must generalize the concept of tangent spaces of Euclidean spaces, that is, it must be an extension of $T_{\rm canonical}.$ In fact this extension should be natural, that is to say, the restriction of $T$ to $\mathcal M(\mathbb E)$ is nutural isomorphic to $T_{\rm canonical}.$ In other words, for any object $(x,\mathbb R^n)$ there is an isomorphism $\alpha_{(x,\mathbb R^m)}:T(x,\mathbb R^n)\to T_{\rm canonical}(x,\mathbb R^n)=\mathbb R^n$ such that if $(x,f,\mathbb R^n,\mathbb R^m):(x,\mathbb R^n)\to (y,\mathbb R^m)$ is a morphism, then the following diagram commutes: enter image description here We say that a functor $T:\mathcal M\to\mathsf{Vec}(\mathbb R)$ gives a definition of tangent spaces, if it satisfies the followng two conditions:

  1. $T$ is a natural extension of $T_{\rm canonical}.$
  2. (Principle of localization)If $(x,f,M,N),(x,g,M,N)$ are morphisms and $f,g$ coincide on a neighborhood of $x$ then $T(x,f,M,N)=T(x,g,M,N).$

Of course, if $T$ and $\widetilde T$ are two such functors, then their restriction to $\mathcal M(\mathbb E)$ are naturally isomorphic through the transformation $\beta:=\widetilde\alpha^{-1}\circ\alpha,$ since we have the commutative diagram enter image description here

To show the uniqueness of tangent space functor, we need to find a natural isomorphism between $T$ and $\widetilde T.$

For any object $(x,M)$ of $\mathcal M,$ we define a vector space isomorphism $\gamma_{(x,M)}:T(x,M)\to \widetilde T(x,M)$ by \begin{equation}\boxed{\gamma_{(x,M)}:=\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)}\end{equation} in which $(U,\varphi)$ is a coordinate chart for $x$ in $M.$ To show that $\gamma_{(x,M)}$ is well-defined, we have to verify that if $(V,\psi)$ is another coordinate chart for $x$ in $M$ then \begin{equation}\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(x),\psi^{-1},\mathbb R^n,M)\circ \beta_{(\psi(x),\mathbb R^n)}\circ T(x,\psi,M,\mathbb R^n)\end{equation}

In fact, principle of localization yields the commutative diagram enter image description here

A similar digram commutes for the functor $\widetilde T.$ On the other hand, since $\psi\circ\varphi^{-1}$ is a functional relation on $\mathbb R^n\times\mathbb R^n,$ it follows that the following diagram commutes: enter image description here

Thus we have a combined commutative diagram

enter image description here

which is desired.

It remains to show that $\gamma$ is a natural isomorphism between $T$ and $\widetilde T.$ Let $(x,f,M,N)$ be a morphism from $(x,M)$ to $(y,N)$ and let $(U,\varphi)$ and $(V,\psi)$ be a coordinate chart for $x$ and $y,$ respectively, then $(\varphi(x),\psi\circ f\circ\varphi,\mathbb R^n,\mathbb R^n)$ is a morphism from $(\varphi(x),\mathbb R^n)$ to $(\psi(y),\mathbb R^n),$ hence \begin{equation}\label{key}\widetilde T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\circ\beta_{\varphi(x),\mathbb R^n}=\beta_{\psi(y),\mathbb R^n}\circ T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\end{equation} Plugging \begin{equation}(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)=(y,\psi,N,\mathbb R^n)\circ (x,f,M,N)\circ(\varphi(x),\varphi^{-1},\mathbb R^n,M)\end{equation} into the former equation gives \begin{equation}\widetilde T(x,f,M,N)\circ \widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(y),\psi^{-1},\mathbb R^n,N)\circ\beta_{(\psi(y),\mathbb R^n)}\circ T(y,\psi,N,\mathbb R^n)\circ T(x,f,M,N)\end{equation} hence we have the commutative diagram enter image description here as expected.

Remark: The principle of localization is equivalent to the following statement: for any two open subsets $U,U'$ of any manifold $M,$ if $x\in U\cap U'$ then there is an isomorphism $\Delta^{(x,M)}_{U,U'}$(abbreviated as $\Delta_{U,U'}$) satisfying the cocycle condition $\Delta_{U',U''}\circ \Delta_{U,U'}=\Delta_{U,U''},\ \Delta_{U,U}=\mathrm{id}$ and if $(x,f,M,N)$ is a morphism from $(x,M)$ to $(y,N),$ then the following diagram commutes: enter image description here Here $f|_{U}^V:=f\cap (U\times V)$ and it's easy to see that $(x,f|_U^V,U,V)$ is a morphism from $(x,U)$ to $(y,V).$

In fact, assume that the principle of localization hold, then defining \begin{equation}\Delta_{U,U'}^{(x,M)}:=T(x,\mathrm{id}_{U\cap U'},U,U')\end{equation} will make the cocycle diagram commute.

Conversely,assume that the cocycle diagram commute, and assume $f$ coincide $g$ on a neighborhood $U$ of $x,$ then we have the commutative diagram enter image description here Since $\Delta_{U,M}\circ\Delta_{M,U}=\mathrm{id},\ \Delta_{N,N}=\mathrm{id},$ it follows that $T(x,f,M,N)=T(x,g,M,N),$ which establishes the equivalence of the two statements.


Here's an approach that I find just abstract enough for my taste. It's kind of an elaboration of p.4 of @anomaly's answer and I found it in (a Soviet translation of) a book by Godbillon [1].

  1. The russians have a very nice term for a fiber bundle $E\to_p B$ which literally translates like a "locally trivial layer'ization". I stress this because it resonates quite well with the formal definition which says that every $b\in B$ has a neighbourhood $U$ whose preimage (under the projection $p$) is homeomorphic to the trivial "layer'ization" $U\times F$ (i.e. just a product). If you repeat these words enough times, the meaning behind them starts really to sink.

  2. Given a topological space $B=\bigcup_\alpha U_\alpha$ covered by open "patches" $U_\alpha$, a theorem states that to turn it into a vector bundle (a locally trivial "layer'ization" with layers (fibers) $F$) it is enough to provide "glue" $g_{\alpha\beta}$ for the patch intersections $U_\alpha\cap U_\beta$ (when non-empty).
    When can trivially layer'ize each patch to get $U_\alpha \times F$ and $U_\beta \times F$, but to glue them together, we have to tell how the layers provided by the two trivializations (at the intersection points $x\in U_\alpha\cap U_\beta$) correspond to each other: so we need a function $g_{\alpha\beta}(x): F\to F$. In the vector bundle case it has to be linear, so $g_{\alpha\beta}(x): \textrm{GL}(F)$. The term for it is the structure group.

  3. Note that the above two points did not mention manifolds at all. A smooth manifold is by definition a union of patches $U_\alpha$ homeomorphic to subsets of $\mathbb R^n$ such that $\varphi_\beta\circ\varphi_\alpha^{-1}$ (maps between subsets of $\mathbb R^n$) are continuously differentiable. Now it is natural as hell to turn it into a vector bundle without ever thinking about tangent stuff.
    This definition matches directly the requirement of the theorem from point 2: we have a topological space covered by patches $M=\bigcup_\alpha U_\alpha$ and the differentiability condition of $\varphi_\beta\circ\varphi_\alpha^{-1}$ provides us a linear map - the derivative - at each intersection point, which we can take as the glue: $g_{\alpha\beta}(x)=D(\varphi_\beta\circ\varphi_\alpha^{-1})|_x$.
    Of course some properties have to be checked before applying the theorem, but in the end we get a vector bundle $TM$ with base space $M$ and fibers homeomorphic to $\mathbb R^n$. It's kind of like it exists without us ever constructing derivations, equivalence classes of curves, etc, right from the start, solely from the definition of a manifold and fiber bundle theory.

  4. In that sense I find this fiber bundle abstract. To get back on the ground, we have to see what's "tangential" about it.

    • First, around each point of the manifold is a neighbourhood $U$ whose $p^{-1}(U)$ ($p$ being the projection $TM\to M$) is homeomorphic to $U\times \mathbb R^n$ - that makes sense because "flat" spaces have precisely $\mathbb R^n$ as tangents.

    • Second, by another vector bundle theory theorem, a continuous map $h:B\to B'$ induces a fiber bundle homomorphism $H:E\to E'$, provided (like above) some glue $h_{\gamma\alpha}(x)\in\textrm{Hom}(F,F')$ is given at each $x\in h^{-1}(U_\gamma ')\cap U_\alpha\subseteq B$ (if non-empty, for each pair of charts).
      In the manifold case, a smooth map $f:M\to N$ induces a map $Tf:TM\to TN$ where the glue is provided by (like above) the derivative of $h$, considered as an $\mathbb R^m\to\mathbb R^n$ function: $h_{\gamma\alpha}(x)=D(\varphi_\gamma'\circ h\circ\varphi_\alpha^{-1})|_x$.

    • To get hold of a tangent vector given a differentiable curve $c:I\to M$ now it natural to lift the curve to $Tc:TI\to TM$. Since $TI=I \times \mathbb R$ is the (truly!) trivial layer'ization, $Tc$ is a function of two argumens, and by that, linear in the second one. So at any point $c(t)$, $Tc(t,\cdot):\mathbb R\to T_{c(t)}M$ is linear (note that the target set - the fiber - is homeomorphic to $\mathbb R^n$ and hence has an induced linear structure) and we can put in the arbitrary number $1$ to get a tangent vector $Tc(t,1)$!

    • I haven't bothered to check if/why every element of $T_{c(t)}M$ is generated by a smooth curve in this way, but I feel that this strategy should work: take a neighbourhood $U$ of $x$ that is both homeomorphic to a subset of $\mathbb R^n$ (by manifoldness of $M$) and which is fiber-trivializable (by vector-bundledness of $TM$). This will get you in the drawing space $U\times \mathbb R^n\cong \mathbb R^n \times \mathbb R^n$. Draw some curve tangential to the desired vector, and bring the curve back to $M$ (by the homeomorphisms used to get in the drawing space).

  5. It's time to stop but let me share one thing more. The gluing of the fibers idea extends even more: if you have a way to transform $\mathrm{GL}(F)$ into $\mathrm{GL}(F')$ (and do it continuously), then you automatically get the so called Associated bundle - a bundle over the same base space $B$, but with fibers $F'$. The idea is as in p.2: the original fiber bundle was constructed from local trivializations $U_\alpha\times F$, with fibers glued at intersections $U_\alpha\cap U_\beta$ by $g_{\alpha\beta}\in \mathrm{GL}(F)$. We can harshly replace these charts with trivializations $U\times F'$ (no knowledge about $F,F'$s whatsoever), but we now have to provide a glue for the new fibers. So if we can get $GL(F')$ from $GL(F)$, we have "shaked up" the old glue with new ingredients and we get the mentioned associated bundle.

    • For example, $g\in\mathrm{GL}(F)$ can be easily converted to $g'\in\mathrm{GL}(F^*)$ by $g'(f^*)=(v\mapsto f^*(gv))$, thus we get the same bundle but with fibers $F^*$ instead of $F$. We can do wedge products $F\wedge F$ and $F^* \wedge F^*$ easily too.
    • Note that this did not have anything to do with manifolds either. But applying it to $TM$, we get the cotangent space $T^*M$ and wedge-product-bundles "for free", purely from fiber bundle theory.

[1] The book I read about this approach is "Differential geometry and analytical mechanics" by Claude Godbillon, but I couldn't find any English translation. Here's an amazon page of the original in French https://www.amazon.fr/G%C3%A9om%C3%A9trie-diff%C3%A9rentielle-m%C3%A9canique-analytique-Godbillon/dp/2705656588


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .